Derivative

Browse the course topics below.

Slope

Introduction

Derivative

Introduction and Geometry

Derivative

Techniques

Applications

Tangent and Normal line

Applications

Increasing and Decreasing

Application

Nature of Points

Application

Concavity

Application

Maxima/Minima

Application

Approximation

Test

Examples

Questions

Old and New

Introduction to Slope

सामान्यतयाः

gradient =\( \frac{\text{change in y}}{\text{change in x}} \)

भन्ने बुझिन्छ। जसलाई तलको चित्रबाट हेरौ।

गणितमा slope ले सीधा रेखाको ठाडोपन (वा तेस्रोपन वा बाङ्गोपन) को मापन गर्दछ। सीधा रेखाको slope गणना गर्न हामी कुनै पनि दुई बिन्दुहरू लिन्छौ र पहिलो बिन्दुबाट दोस्रो बिन्दुमा run र rise को गणना गर्छौ। Run भनेको x-निर्देशांकहरूमा हुने परिवर्तन हो र rise भनेको y-निर्देशांकहरूमा भएको परिवर्तन हो, जसलाई तलको चित्र [चित्र 1] मा चित्रण गरिएको छ।

त्यसैले,
slope =\( \frac{rise}{run} \) (1)
सीधा रेखाको slope धनात्मक, ऋणात्मक वा शून्य हुन सक्छ, जुन कुरा पहिलो बिन्दुबाट दोस्रो बिन्दुमा coordinate हरुको मान बढ्छ, घट्छ वा उस्तै रहन्छ भन्नेमा निर्भर गर्दछ।
खास गरी, बायाँबाट दायाँ जादा रेखाको स्थिति तल गएमा ऋणात्मक slope, तेर्सो भएमा शून्य slope हुन्छ, र माथि गएमा धनात्मक slope हुन्छ ।
यदि Run को मान शून्य छ भने रेखाको slope अपरिभाषित हुन्छ, किनकि शून्यले भाग गर्ने कुरा सम्भव छैन। त्यसैले ठाडो रेखाहरूको slope अपरिभाषित हुन्छन्।
गणितिय रुपमा, दुईवटा बिन्दुहरू \( (x_1, y_1)\) र \( (x_2, y_2)\) छन भने ति बिन्दुहरुले बनाउने सिधा रेखाको slope तलको चित्रमा चित्रण गरे जस्तै (चित्र 2) हुन्छ।

Fig.2

Then
run = x₂ − x₁ and rise = y₂ − y₁.
Now,
gradient =\( \frac{y_2-y_1}{x_2-x_1} \) (2)
Remember that when we use this formula to calculate the gradient of a straight line, it doesn’t matter which point we take to be the first point, either \( (x_1, y_1)\) or \( (x_2, y_2)\), we get the same result either way.

Example 1

Calculate the gradient of a line passing through the points (1, 8) and (5,2).

Fig.3

The slope is
gradient =\( \frac{y_2-y_1}{x_2-x_1}=\frac{2-8}{5-1}=-\frac{3}{2} \)
OR
gradient =\( \frac{y_2-y_1}{x_2-x_1}=\frac{8-2}{1-5}=-\frac{3}{2} \)

Introduction to Derivative

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Introduction

The derivative is a fundamental tool of calculus that quantifies the sensitivity of change of a function.

1. Gottfried Wilhelm Leibniz in 1675 has given a common symbol for the derivative of a function as \( \displaystyle \frac{dy}{dx}\).
   Calculus को आधारभूत उपकरण मा derivative पनि पर्दछ। जसले sensitivity of change of a function को मापन गर्दछ। 1675 मा Gottfried Wilhelm Leibniz ले यसलाई\( \displaystyle \frac{dy}{dx}\) को संकेतले जनाएका थिए।
2. Joseph-Louis Lagrange has given another common notation for differentiation by using the prime mark in the symbol of a function \( \displaystyle f'(x)\) or \(y'\).
   Joseph-Louis Lagrange ले derivativeलाई \( \displaystyle f'(x)\) or \(y'\) prime notation बाट जनाएका थिए।
3. Newton ले derivativeलाई dot notation ले जनाएका थिए। This notation is used exclusively for derivatives with respect to time or arc length. It is typically used in differential equations in physics and differential geometry.However, the dot notation becomes unmanageable for high-order derivatives (of order 4 or more) and cannot deal with multiple independent variables.
4. Another notation for derivative is D-notation. The first derivative is written Df(x) and higher derivatives are written with a superscript \(D^{n}f(x)\). This notation is sometimes called Euler notation.
   To indicate a partial derivative, the variable differentiated by is indicated with a subscript, for example given the function u=f(x,y),} its partial derivative with respect to x can be written \( D_{x}u\) or \( D_{x}f(x,y)\)

In mathematics, derivative is defined as the method that shows the simultaneous rate of change. It can be examplified by using slope and limit.

Test your Understanding: Quiz 2

1. Which of the following is the derivative of a function \(f(x)\) at a point \(x=a\)?
     \(\displaystyle \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \)
     \( \frac{f(a+h) - f(a)}{h} \)
     \(\displaystyle \lim_{h \to 0} \frac{f(a+h) + f(a)}{h}\)
     \( \frac{f(a+h) - f(a-h)}{h} \)
2. Which of the following conditions is necessary for a function \(f(x)\) to be differentiable at a point \(x=a\)?
     f(x) must have a vertical tangent at \(x=a\)
     f(x) must have a horizontal tangent at \(x=a\)
     f(x) must be continuous at \(x=a\)
     f(x) must have a limit at \(x=a\)
3. Which is NOT a derivative of \(y=f(x)\)?
     \( f'(x) \)
     \( \frac{dy}{dx} \)
     \( y \Delta x \)
     \( D_x y \)
4. Which of the following line's slope is represented by \(f'(x)\)?
     Secant
     Normal
     Asymptote
     Tangent
5. At which feature on the graph of a continuous function \(f(x)\) is the derivative \(f'(x)\) most likely to be undefined?
     A smooth curve at a local maximum
     An inflection point
     A corner or a cusp
     A horizontal tangent line
6. The height of a tree \(H\) (in meters) is a function of its age \(A\) (in years), \(H(A)\). What are the units of the derivative \(H'(A)\)?
     Years/meter
     Meters·years
     Meters/year
     Unitless

Derivative as Instantaneous Rate of Change

This is the rate of change of a quantity at a specific instant in time or a specific point in space. In calculus, the derivative of a function represents the rate at which the function is changing at any given point.

In the figure above, if we need the change in function at an instant point , say x=a, then the instantaneous rate of change at x=a is given by
change=\(\frac{f(a+\delta)-f(a-\delta)}{2 \delta}\)

Example

1. Measure the change in function \(f(x)=x^2\) at an instant point , say at x=2.
   Solution
   The instantaneous rate of change at x=2 is
   instantaneous rate of change=\(\frac{f(2+0.1)-f(2-0.1)}{2 \times 0.1}=\frac{f(2.1)-f(1.9)}{0.2}=\frac{2.1^2-1.9^2}{0.2}=4\)
2. An object falling from rest has displacement s in cm given by \(s = 490t^2\), where t is in seconds (s).What is the velocity when t = 10s?
   Solution
   The instantaneous rate of change at t = 10s is
   instantaneous rate of change=\(\frac{f(10+0.1)-f(10-0.1)}{2 \times 0.1}=\frac{f(10.1)-f(9.9)}{0.2}=\frac{10.1^2-9.9^2}{0.2}=98m/s\)

Test your Understanding: Quiz 3

1. If \(s(t)\) represents the position of an object at time \(t\), what does the value of \(s'(5)\) represent?
     The object's initial velocity
     The object's total displacement after 5 units of time
     The average velocity over the first 5 units of time
     The instantaneous velocity at time \(t=5\)
2. A balloon is being inflated such that its volume \(V\) is a function of time \(t\). If \(\frac{dV}{dt} = 10 \text{ cm}^3/\text{min}\), what does this derivative represent?
     The total change in volume over 10 minutes
     The instantaneous rate of change in volume
     The average rate of change in volume
     Total change in area of the balloon
3. The instantaneous rate of change of a function \(f(x)\) is generally a better measure of change at a point \(x=c\) than the average rate of change over \([c, c+h]\) because:
     The instantaneous rate is always a larger number
     It takes the limit as the interval approaches infinity
     It takes the limit as the interval approaches 0
     It is easier to calculate
4. The process of local linear approximation uses the derivative to approximate the function's value near a known point. This is based on the idea that every differentiable function looks like a \( \dots \) when viewed closely enough.
     Hyperbola
     Parabola
     Straight line
     Circle
5. The formula for the linear approximation of \(f(x)\) at \(x=a\) is \(f(a) + f'(a)(x-a)\). This is mathematically the same as the equation for the \( \dots \)
     secant
     tangent
     normal
     quadratic approximation
6. Given \(f(3)=10\) and \(f'(3)=2\). Use the linear approximation to estimate \(f(3.5)\).
     \(10.5\)
     \(11\)
     \(12\)
     \(11.5\)
7. Which is the differential used to approximate the change in \(f(x) = x^2\) as \(x\) changes from \(3\) to \(3.01\)?
     \(0.01\)
     \(0.06\)
     \(0.09\)
     \(0.005\)

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Derivative as Slope of tangent

We define derivative as slope of tangent. Please remember, the word ‘tangent’ comes from the Latin word ‘tangere’, which means ‘to touch’. The English word ‘tangible’, which means ‘capable of being touched’, comes from the same Latin word.

Let \(f (x)\) be a differentiable function at x.
Let \(P(x, f(x))\) and \(Q(x + h, f(x+h))\) be two nearby points.
Then PQ is a secant.
Now, the slope of the secant PQ is
slope of secant at \( P = \frac{f(x+h)-f(x) }{x + h- x}\)
or slope of secant at \( P = \frac{ f(x+h)-f(x) }{h}\) (A)
This expression (A) is known as the difference quotient for the function \(f(x)\) at x
Taking limit as \( h \to 0\) in (A), the secant will turn into tangent at P ,
Then, the slope of tangent at P is
slope of tangent at \( P = \displaystyle \lim_{h \to 0} \frac{f(x+h)-f(x)}{ h}\)
orslope of tangent at P = \(f'(x)\)
This new function \(f'(x)\) is called the derivative (or derived function) of the function \(f(x)\).
The process of finding the derivative of a given function f(x) is called differentiation
For example, a function is \( f(x) = x^2\), and its derivative is f'(x) = 2x.

Test your Understanding: Quiz 4

1. Which statement accurately describes the geometric meaning of \(f'(a) = 5\)?
     The function's value is 5 at \(x=a\).
     The tangent line to \(f(x)\) at \(x=a\) has a slope of 5.
     The average rate of change on \([a, a+1]\) is 5.
     The function has a local maximum at \(x=5\).
2. If the tangent line to \(f(x)\) at \(x=a\) is horizontal, what must be \(f'(a)\)?
     Undefined
     1
     The value of \(f(a)\)
     0
3. The derivative of \(f(x)\) is \(f'(x) = x - 3\). At what \(x\)-value is the tangent line horizontal?
     \(x=3\)
     \(x=0\)
     \(x=-3\)
     Nowhere
4. The slope of the tangent line to \(f(x) = x^4 - 2x^2 + 5\) at \(x = 1\) is:
5. What is the slope of the tangent line to \(y = \frac{3}{x}\) at \(x=-1\)?
6. The slope of the tangent line to \(g(t) = t^3 + 5t\) at \(t=2\) is:
7. Find the slope of the tangent line to \(h(x) = 2\sqrt{x}\) at \(x=9\):
8. What is the slope of the tangent line to \(f(x) = (2x - 1)^2\) at \(x=1\)?

Derivative using Definition/First Principle

Example 1

Use the definition of the derivative to compute the derivative of \( f (x)\)
Use the following process
step1: \( f (x)\)
step2: \( f (x+h)\)
step3: \( f (x+h)-f (x)\)
step4: \( \frac{f (x+h)-f (x)}{h}\)
step5: Simplify \( \frac{f (x+h)-f (x)}{h}\)
step6: \( \displaystyle \lim_{h \to 0} \frac{f (x+h)-f (x)}{h}\)

Example 2

Use the definition of the derivative to compute the derivative of \( f (x)= x^n\)
The solution is
\( f'(x)=\displaystyle \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\)
or \( f'(x)=\displaystyle \lim_{h\to 0}\frac{{{( x+h)}^{n}}-{x^n}}{h}\)
or \( f'(x)=\displaystyle \lim_{h\to 0}\frac{( {x^n}+n{x^{n-1}}h+...+xh^{n-1}+h^n)-{x^n}}{h}\)
or \( f'(x)=\displaystyle \lim_{h\to 0}\frac{n{x^{n-1}}h+...+xhx^{n-1}+h^n}{h}\)
or \( f'(x)=\displaystyle \lim_{h\to 0}\frac{h( n{x^{n-1}}+...+xh^{n-2}+h^{n-1})}{h}\)
or \( f'(x)=\displaystyle \lim_{h\to 0}( n{x^{n-1}}+...+xh^{n-2}+h^{n-1})\)
or \( f'(x)=n{x^{n-1}}\)

Example 3

Find the derivative of \( f(x)=\sin x\) by first principle.
By definition, the derivative of \( f(x)\) is
\( f'(x)=\displaystyle \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\)
or \( f'(x)=\displaystyle \lim_{h\to 0}\frac{\sin ( x+h)-\sin x}{h}\)
or \( f'(x)=\displaystyle \lim_{h\to 0}\frac{2\cos ( \frac{x+h+x}{2}).\sin ( \frac{x+h-x}{2})}{h}\)
or \( f'(x)=\displaystyle \lim_{h\to 0}\frac{2\cos ( x+\frac{h}{2}).\sin ( \frac{h}{2})}{h}\)
or \( f'(x)=\displaystyle \lim_{h\to 0}\cos ( x+\frac{h}{2}).\frac{\sin ( \frac{h}{2})}{\frac{h}{2}}\)
or \( f'(x)=\displaystyle \lim_{h\to 0}\cos ( x+\frac{h}{2}).\displaystyle \lim_{h\to 0}\frac{\sin ( \frac{h}{2})}{\frac{h}{2}}\)
or \( f'(x)=\cos ( x+\frac{0}{2}).1\)
or \( f'(x)=\cos x \)

Example 4

Find the derivative of \( f(x)=e^x \) by first principle.
By definition, the derivative of \( f(x)\) is
\( f'(x)=\displaystyle \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\)
or \( f'(x)=\displaystyle \lim_{h\to 0}\frac{e^{x+h} -e^x }{h}\)
or \( f'(x)=\displaystyle \lim_{h\to 0}\frac{e^x ( e^{h} -1)}{h}\)
or \( f'(x)=e^x \displaystyle \lim_{h\to 0}\frac{e^{h} -1}{h}\)
or \( f'(x)=e^x \)

Example 5

Find the Derivative of \( f(x)=\log x\) by first principle.
By definition, the derivative of \( f(x)\) is
\( f'(x)=\displaystyle \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\)
or \( f'(x)=\displaystyle \lim_{h\to 0}\frac{\log (x+h)-\log x}{h}\)
or \( f'(x)=\displaystyle \lim_{h\to 0}\frac{\log ( \frac{x+h}{x})}{h}\)
or \( f'(x)=\displaystyle \lim_{h\to 0}\frac{\log ( 1+\frac{h}{x})}{h}\)
or \( f'(x)=\displaystyle \lim_{h\to 0}\frac{1}{h}\log ( 1+\frac{h}{x})\)
or \( f'(x)=\displaystyle \lim_{h\to 0}\log {{( 1+\frac{h}{x})}^{\frac{1}{h}}}\)
or \( f'(x)=\log \{ \displaystyle \lim_{h\to 0}{{( 1+\frac{h}{x})}^{\frac{1}{h}}}\}\)
Let \( u=\frac{h}{x}\) then, \( \frac{1}{h}=\frac{1}{ux}\) , and \( h\to 0\) implies \( u\to 0\)
\( f'(x)=\log \{ \displaystyle \lim_{u\to 0}{{( 1+u)}^{\frac{1}{ux}}}\}\)
or \( f'(x)=\log {{ \left \{ \displaystyle \lim_{u\to 0}{{( 1+u)}^{\frac{1}{u}}} \right \}}^{\frac{1}{x}}}\)
or \( f'(x)=\log { {\left \{ e \right \}}^{\frac{1}{x}}}\)
or \( f'(x)=\frac{1}{x}\)

Example 6

Use the definition of the derivative to compute the derivative of \( f (x)= x +\frac{1}{x}\)
Use the following process
step1: \( f (x)= x +\frac{1}{x}\)
step2: \( f (x+h)= x+h +\frac{1}{x+h}\)
step3:\( f (x+h)-f (x)= x+h +\frac{1}{x+h}-x-\frac{1}{x}\)
step4:\( \frac{f (x+h)-f (x)}{h}= \frac{x+h +\frac{1}{x+h}-x-\frac{1}{x}}{h}\)
step5:Simplify \( \frac{f (x+h)-f (x)}{h}=\frac{h +\frac{1}{x+h}-\frac{1}{x}}{h}= \frac{ x^2+ h x-1 }{x (h + x)}\)
step6: \( \displaystyle \lim_{h \to 0} \frac{f (x+h)-f (x)}{h}= 1 - \frac{1}{x^2}\)

Derivative Techniques

Technique	When to Use	Key Formula
Power Rule	Monomials \( x^n \)	\( nx^{n-1} \)
Constant Multiple	Constant \(\times\) function	\( c f'(x) \)
Sum/Difference	Add/subtract functions	\( f' \pm g' \)
Product Rule	Product of two functions	\( u'v + uv' \)
Quotient Rule	Quotient of two functions	\( \dfrac{u'v - uv'}{v^2} \)
Chain Rule	Composite functions	\( f'(g(x)) g'(x) \)
Implicit Diff.	\( y \) not isolated	Differentiate & solve for \( y' \)
Inverse Functions	\( f^{-1}(x) \)	\( \dfrac{1}{f'(f^{-1}(x))} \)
Logarithmic Diff.	Variable exponents	Take \(\log\), then differentiate
Exponential Diff.
Higher-Order	Repeated differentiation	\( f^{(n)}(x) \)

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1. Power Rule

Rule: If \( f(x) = x^n \), where \( n \in \mathbb{R} \), then

\[ f'(x) = n x^{n-1} \]

Solution 👉 Click Here

Proof
We know that \begin{align*} \frac{df}{dx} &= \lim_{h\to 0}\frac{f(x+h)-f(x)}{h} \\ &=\lim_{h\to 0}\frac{{{( x+h)}^{n}}-{x^n}}{h}\\ &= \lim_{h\to 0}\frac{( {x^n}+n{x^{n-1}}h+...+xh^{n-1}+h^n)-{x^n}}{h}\\ &=\lim_{h\to 0}\frac{n{x^{n-1}}h+...+xhx^{n-1}+h^n}{h}\\ &= \lim_{h\to 0}\frac{h( n{x^{n-1}}+...+xh^{n-2}+h^{n-1})}{h}\\ &=\lim_{h\to 0}( n{x^{n-1}}+...+xh^{n-2}+h^{n-1})\\ &=n{x^{n-1}} \end{align*}

For example: The derivative of \( f(x) = x^5 \) is

\[ f'(x) = 5x^4 \]

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2. Constant Multiple Rule

Rule: If \( f(x) = c \cdot g(x) \), \( c \) constant, then

\[ f'(x) = c \cdot g'(x) \]

Solution 👉 Click Here

Proof
We know that \begin{align*} \frac{df}{dx} &= \lim_{h \to 0} \frac{c\,g(x+h) - c\,g(x)}{h} \\ &= \lim_{h \to 0} \frac{c\,\big( g(x+h) - g(x) \big)}{h} \\ &= c\,\lim_{h \to 0} \frac{g(x+h) - g(x)}{h} \\ &= c\,g'(x) \end{align*}

For example: The derivative of \( f(x) = 7x^3 \) is

\[ f'(x) = 7 \cdot 3x^2 = 21x^2 \]

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3. Sum and Difference Rule

Rule: If \( f(x) = g(x) \pm h(x) \), then

\[ f'(x) = g'(x) \pm h'(x) \]

Solution 👉 Click Here

Proof
We know that \begin{align*} \frac{df}{dx} &= \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \\ &= \lim_{h \to 0} \frac{g(x+h) + h(x+h) - g(x) - h(x)}{h} \\ &= \lim_{h \to 0} \frac{g(x+h) - g(x) + h(x+h) - h(x)}{h} \\ &= \lim_{h \to 0} \left( \frac{g(x+h) - g(x)}{h} + \frac{h(x+h) - h(x)}{h} \right) \\ &= \lim_{h \to 0} \frac{g(x+h) - g(x)}{h} + \lim_{h \to 0} \frac{h(x+h) - h(x)}{h}\\ &= g'(x)+h\(x) \end{align*}

For example: The derivative of\( f(x) = 4x^2 - 3x + 1 \) is

\[ f'(x) = 8x - 3 \]

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4. Product Rule

Rule: If \( f(x) = u(x) \cdot v(x) \), then

\[ f'(x) = u'(x)v(x) + u(x)v'(x) \]

Solution 👉 Click Here

Proof
We know that \begin{align*} \frac{d}{dx}[g(x)h(x)] &= \lim_{k \to 0} \frac{g(x+k)h(x+k) - g(x)h(x)}{k} \\ &= \lim_{k \to 0} \frac{g(x+k)h(x+k) - g(x)h(x+k) + g(x)h(x+k) - g(x)h(x)}{k} \\ &= \lim_{k \to 0} \left[ \frac{g(x+k) - g(x)}{k} \, h(x+k) + g(x) \, \frac{h(x+k) - h(x)}{k} \right] \\ &= \left( \lim_{k \to 0} \frac{g(x+k) - g(x)}{k} \right) \left( \lim_{k \to 0} h(x+k) \right) + g(x) \left( \lim_{k \to 0} \frac{h(x+k) - h(x)}{k} \right) \\ &= g'(x) \, h(x) + g(x) \, h'(x) \end{align*}

For example: The derivative of \( f(x) = x^2 \sin x \) is

\[ f'(x) = 2x \sin x + x^2 \cos x \]

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5. Quotient Rule

Rule: If \( f(x) = \dfrac{u(x)}{v(x)} \) with \( v(x) \ne 0 \), then

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} \]

Solution 👉 Click Here

Proof
We know that \begin{align*} \frac{d}{dx}\!\left[\frac{g(x)}{h(x)}\right] &= \lim_{k \to 0} \frac{\frac{g(x+k)}{h(x+k)} - \frac{g(x)}{h(x)}}{k} \\[6pt] &= \lim_{k \to 0} \frac{g(x+k)h(x) - g(x)h(x+k)}{k\,h(x+k)h(x)} \\[6pt] &= \lim_{k \to 0} \frac{g(x+k)h(x) - g(x)h(x) + g(x)h(x) - g(x)h(x+k)}{k\,h(x+k)h(x)} \\[6pt] &= \lim_{k \to 0} \frac{ \big(g(x+k) - g(x)\big) h(x) - g(x) \big( h(x+k) - h(x) \big) }{k\,h(x+k)h(x)} \\[6pt] &= \lim_{k \to 0} \left[ \frac{ \frac{g(x+k) - g(x)}{k} \, h(x) - g(x) \, \frac{h(x+k) - h(x)}{k} } { h(x+k) h(x) } \right] \\[6pt] &= \frac{ g'(x) h(x) - g(x) h'(x) }{ [h(x)]^2 } \end{align*}

For example: The derivative of\( f(x) = \dfrac{x}{x^2 + 1} \) is

\[ f'(x) = \frac{(1)(x^2 + 1) - x(2x)}{(x^2 + 1)^2} = \frac{1 - x^2}{(x^2 + 1)^2} \]

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6. Chain Rule

Rule: If \( y = f(g(x)) \), then

\[ \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \]

Solution 👉 Click Here

Proof
We know that \begin{align*} \frac{d}{dx}\big[g(h(x))\big] &= \lim_{k \to 0} \frac{g(h(x+k)) - g(h(x))}{k} \\[6pt] &= \lim_{k \to 0} \left( \frac{g(h(x+k)) - g(h(x))}{h(x+k) - h(x)} \cdot \frac{h(x+k) - h(x)}{k} \right) \\[6pt] &= \left( \lim_{k \to 0} \frac{g(h(x+k)) - g(h(x))}{h(x+k) - h(x)} \right) \cdot \left( \lim_{k \to 0} \frac{h(x+k) - h(x)}{k} \right) \\[6pt] &= g'\big(h(x)\big) \cdot h'(x) \end{align*}

For example: The derivative of\( f(x) = \sin(3x^2) \) is

\[ f'(x) = \cos(3x^2) \cdot 6x = 6x \cos(3x^2) \]

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7. Implicit Differentiation

When: \( y \) is defined implicitly (e.g., \( x^2 + y^2 = 25 \)), then
Differentiate both sides w.r.t. \( x \), then we get

\[ \dfrac{dy}{dx} \]

For example, to find \( \dfrac{dy}{dx} \) for \( x^2 + y^2 = 25 \), we do \[x^2 + y^2 = 25\] \[2x + 2y \frac{dy}{dx} = 0\] \[\frac{dy}{dx} = -\frac{x}{y}\]

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8. Derivatives of Inverse Functions

Let \( y = f^{-1}(x) \), then we can write \( x = f(y) \) and differentiate both sides implicitly with respect to \( x \) , we get \[ 1 = f'(y)\cdot \frac{dy}{dx}. \] Thus, \[ \frac{dy}{dx} = \frac{1}{f'(y)} \] but \( y = f^{-1}(x) \), thus \[ \left[f^{-1}\right]'(x) = \frac{1}{f'\!\left(f^{-1}(x)\right)}. \] For example
Find the derivative of \( \sin^{-1}(x) \).
As above, we write \( y = \sin^{-1}(x) \), so \( x = \sin(y) \) \[ \frac{d}{dx}x = \frac{d}{dx}\sin(y) \] \[ 1 = \cos(y)\cdot \frac{dy}{dx} \] \[ \frac{dy}{dx} = \frac{1}{\cos y} \] \[ \frac{dy}{dx} = \frac{1}{\sqrt{1-\sin^2y}} \] \[ \frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}} \]

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9. Logarithmic Differentiation

Let \( y = f(x)^{g(x)} \) , then

Take \(\ln\) of both sides, we get
\[\log y= g(x) \log f(x)\] Differentiate implicitly, and get \(y'\)

For example, find the derivative of\( y = x^x \) for \( x > 0 \), then
\[\log y= x \log x\] \[\frac{1}{y} y' = \log x + 1 \] \[y' = y (\log x + 1)\] \[y' = x^x (\log x + 1)\]

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10. Exponential Differentiation

Rule: If \( y = e^x\), then

\[ \frac{dy}{dx} = e^x \]

Solution 👉 Click Here

Proof
We know that \begin{align*} \frac{d}{dx}\big[e^x\big] &= \lim_{h \to 0} = \lim_{h\to 0}\frac{f(x+h)-f(x)}{h} \\ &= \lim_{h\to 0}\frac{e^{x+h} -e^x }{h}\\ &= \lim_{h\to 0}\frac{e^x ( e^{h} -1)}{h}\\ &= e^x \lim_{h\to 0}\frac{e^{h} -1}{h}\\ &= e^x \end{align*}

For example, find the derivative of\( y = e^{3x} \) then
\[ y = e^{3x}\] \[y' = e^{3x}. 3\] \[y' = 3e^{3x}\]

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11. Higher-Order Derivatives

the Higher-Order Derivatives is Repeated differentiation: for example \( f', f'', f^{(n)} \).

If \( f(x) = e^{2x} \), then

\[f'(x) = 2e^{2x}\] and \[f''(x) = 4e^{2x}\]

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Exit Ticket

Differentiate \( f(x) = \dfrac{(x+1)^2}{\sqrt{x}} \) using two different methods.

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Exercise

1. \(\displaystyle \frac{d}{dx}(a^x) = a^x \log a\)
2. \(\displaystyle \frac{d}{dx}(|x|) = \frac{|x|}{x}, \quad x \ne 0\)
3. \(\displaystyle \frac{d}{dx}(\log_e |x|) = \frac{1}{x}, \quad x \ne 0\)
4. \(\displaystyle \frac{d}{dx}(x^x) = x^x (1 + \log x), \quad x > 0\)
5. \(\displaystyle \frac{d}{dx}(\sin x) = \cos x\)
6. \(\displaystyle \frac{d}{dx}(\cos x) = -\sin x\)
7. \(\displaystyle \frac{d}{dx}(\tan x) = \sec^2 x, \quad \left\{ x \ne n\pi + \frac{\pi}{2} \right\}\)
8. \(\displaystyle \frac{d}{dx}(\cot x) = -\csc^2 x, \quad \left\{ x \ne n\pi \right\}\)
9. \(\displaystyle \frac{d}{dx}(\sec x) = \sec x \tan x, \quad \left\{ x \ne n\pi + \frac{\pi}{2} \right\}\)
10. \(\displaystyle \frac{d}{dx}(\csc x) = -\csc x \cot x, \quad \left\{ x \ne n\pi \right\}\)
11. \(\displaystyle \frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1 - x^2}}, \quad -1 < x < 1\)
12. \(\displaystyle \frac{d}{dx}(\cos^{-1} x) = \frac{-1}{\sqrt{1 - x^2}}, \quad -1 < x < 1\)
13. \(\displaystyle \frac{d}{dx}(\tan^{-1} x) = \frac{1}{1 + x^2}, \quad x \in \mathbb{R}\)
14. \(\displaystyle \frac{d}{dx}(\cot^{-1} x) = \frac{-1}{1 + x^2}, \quad x \in \mathbb{R}\)
15. \(\displaystyle \frac{d}{dx}(\sec^{-1} x) = \frac{1}{|x|\sqrt{x^2 - 1}}, \quad |x| > 1\)
16. \(\displaystyle \frac{d}{dx}(\csc^{-1} x) = \frac{-1}{|x|\sqrt{x^2 - 1}}, \quad |x| > 1\)

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Application of Derivative

1. Approximation
2. Rate of Change of Quantities
3. Slope of a Line
4. Slopes of the Tangent and the Normal
5. Equations of the Tangent and the Normal
6. Angle between Two Lines/Curves
7. Length of Tangent, Normal, Subtangent and Subnormal
8. Increasing and Decreasing Functions
9. Shape of graph
10. Maxima and Minima Functions
11. Rolle’s Theorem
12. Mean Value Theorem

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Equation of tangent line

Tangent is limiting position of secant

A tangent is a straight line that touches a curve at one point. The idea is that the tangent and the curve are both going in exact same direction at point of contact. In a space curve, number of lines may pass touching one point, therefore, a precise definition of tangent is given as below.
Let C be a curve.
Also let A and B are two points on C
when, point B is moved toward A then limiting form of secant AB, is called tangent to the curve C at A.

Drag point A	Drag point B

Equation of tangent line

Let y=f(x) be a curve C
Also let \(P(x_1,y_1)\) be a given point on C.
Then
Equation of straight line passing through P is
\( y-y_1=m(x-x_1)\)
Since, the tangent line has slope \(m=\frac{dy}{dx}\), the equation of tangent line at P is
\( y-y_1=\frac{dy}{dx} (x-x_1)\)
or \( y=y_1+\frac{dy}{dx} (x-x_1)\)
This completes the proof

Increasing/decreasing test of a function

Increasing Function

A function \( y = f (x)\) is said to be increasing in an open interval \( (a, b)\) i.e. between the points \( x = a\) and \( x = b\) if
\( \frac{dy}{dx} \) or \( f ' (x)\) is positive
for all values of \( x\) in that interval.
In other words, f(x) is said to be increasing if
\( x_1 < x_2 \Rightarrow f (x_1) < f (x_2)\)
or equivalently
\( x_1 > x_2 \Rightarrow f (x_1) > f (x_2)\)
for all real numbers \( x_1\) and \( x_2\) in \( (a, b)\).

An increasing function can be shown graphically as in the figure below.
The graph of such a function is a curve which goes on rising with the rise in the value of \( x\).

Decreasing Function

A function \( y = f (x)\) is said to be decreasing in an open interval \( (a, b)\) i.e. between the points \( x = a\) and \( x = b\) if
\( \frac{dy}{dx} \) or \( f ' (x)\) is negative
for all values of \( x\) in that interval.
In other words, f(x) is said to be increasing if
\( x_1 < x_2 \Rightarrow f (x_1) > f (x_2)\)
or equivalently
\( x_1 > x_2 \Rightarrow f (x_1) < f (x_2)\)
for all real numbers \( x_1\) and \( x_2\) in \( (a, b)\).

A decreasing function can be shown graphically as in the figure below.
The graph of such a function is a curve which goes on falling with the raise in the value of \( x\).

Example 1

Show that the function \( f(x) = 2x^3 - 3x^2 - 36x \) is increasing on each of the intervals (-∞,-2) and (3,∞), and decreasing on the interval (-2, 3).

Solution
Find the derivative. The derivative is
\( f'(x) = 6x^2 - 6x - 36\) (1)
Factorizing (i) gives
\( f'(x) = 6(x^2 - x - 6) = 6(x + 2)(x - 3)\)

	(-∞,-2)	(-2,3)	(3,∞)
(x+2)	-	+	+
(x-3)	-	-	+
f'(x)	+	-	+
curve	increasing	decreasing	increasing

1. When x is less than -2, the values of x + 2 and x - 3 are both negative, and hence the value of \( f'(x) = 6(x + 2)(x - 3) \) is positive.
   Therefore, by the increasing/decreasing criterion, the function f is increasing on the interval (-∞,-2).
2. When x is in the interval (-2, 3), the value of x + 2 is positive and the value of x - 3 is negative, and hence the value of \( f'(x) = 6(x + 2)(x - 3)\) is negative.
   Therefore, by the increasing/decreasing criterion, the function f is decreasing on the interval (-2, 3).
3. When x is greater than 3, the values of x + 2 and x - 3 are both positive, and hence the value of \( f'(x) = 6(x + 2)(x - 3)\) is also positive.
   Therefore, by the increasing/decreasing criterion, the function f is increasing on the interval (3,∞).

First Derivative Test

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Can you write a function whose derivative does NOT exist a some point.🗨 🙋
🕵📋 Yes, below is a function fiven by f(x)=|x| whose derivative does NOT exist at x=0 .

Given function is
\(f(x)= |x|\)
In the simplified form, this function can be written as
\( f(x)= \begin{cases} -x &\text { for } x < 0 \\ x &\text { for } x>0 \end{cases} \)
The derivative function is given by
\( f'(x)= \begin{cases} -1 &\text { for } x < 0 \\ 1 &\text { for } x>0 \end{cases} \)
So,
At \( x=0^{-}\), the left hand limit is
\( \displaystyle \lim_{x \to 0^{-}} f'(x)= \lim_{x \to 0^{-} } -1= -1\) LHL
At \( x=0^{+}\), the right hand limit is
\( \displaystyle \lim_{x \to 0^{+}} f'(x)= \lim_{x \to 0^{+} } 1= 1\) RHL
Since, LHL≠RHL, so, the derivative of the function f(x)=|x| does NOT exist at x=0.

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Nature of points

There are 6 types of points, they are describes as below.

1. Critical Point

Critical Point

A point where the derivative of a function is either zero or undefined, is called critical points.
In the figure below, A,B,C, and D are critical points in which

1. f'(x)=0 at A
2. f'(x)=0 at B
3. f'(x) is NOT defined at C (f is not differentiable at C)
4. f'(x)=0 at D

Critical points are the points on the graph where the function's rate of change is altered—either a change from increasing to decreasing, in concavity, or in some unpredictable fashion.

A critical point is

1. stationary point if the function changes from increasing/decreasign to decreasing/increasing at that point
2. turning point if the function changes from increasing/decreasign to decreasing/increasing at that point and and is a local minimum/maximum
3. saddle point if the function changes both increasing/decreasign and concavity at that point

NOTE: Maxima and mimina can occurs at critical points where first derivative is undefined.

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Stationary Point

A point where the derivative of a function (the gradient/slope of a graph) is zero is called a stationary point.
In the figure below, A,B,and D are stationary points in which

1. f'(x)=0 at A
2. f'(x)=0 at B
3. f'(x)=0 at D

Note: all stationary points are critical points, but not all critical points are stationary points.

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Turning Point

A point where the derivative of a function is (the gradient/slope of a graph) is zero and function has a local maximum or local minimum (extrema point), is called a turning point.There are two types of turning points: local maximum and local minimum.

1. If f is increasing on the left interval and decreasing on the right interval, then the stationary point is a local maximum
2. If f is decreasing on the left interval and increasing on the right interval, then the stationary point is a local minimum

In the figure below, C is only the turning point[Please note that A and B is not turning point] in which

1. f'(x)=0 at A, but local extrema is NOT defined at A
2. f'(x) is NOT defined at B
3. f'(x)=0 at C, local minima is defined at C

Note: all turning points are stationary points, but not all stationary points are turning points.

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Straight line/Constant function

Some stationary points are neither turning points nor horizontal points of inflection. For example, every point on the graph of the equation \(y = 1\) (see Figure below), or on any horizontal line, is a stationary point that is neither a turning point nor point of inflection. It is a straight line.

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Point of Inflection/ Plateau point

A point where the second derivative of a function changes sign, the graph of the function will switch from concave down to concave up, or vice versa, is called an inflection point. Assuming the second derivative is continuous, it must take a value of zero at any inflection point, although not every point where the second derivative is zero is necessarily a point of inflection.

Mathematically, a point on a function f (x) is said to be a point of inflexion if f''(x) = 0 and f'''(x)≠0.
At this point, the concavity changes from upward to downward or vice-versa.
Therefore the point of inflexion is the transition between concavity of the curves.

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Saddle point (Horizontal point of inflection)

A saddle point or minimax point is a point on a function where the slopes (derivatives) is zero , but which is not a local extremum of the function. Thus, a saddle point is a point which is both a stationary point and a point of inflection. Since it is a point of inflection, it is not a local extremum.

Figure below shows the graph of the function \( f(x) = x^3\) .
The derivative of this function is \( f'(x) = 3x^2\), so the 1st derivative f'(x) =0 when x = 0 is
\(f'(0) = 3 \times 0^2 = 0\).
So the function \( f(x) = x^3\) has a stationary point at x = 0.
However, this stationary point isn’t a turning point.
Because
The second derivative of this function \( f''(x) = 6x\), so the second derivative \( f''(x) = 0\) when x = 0 is
\(f''(0) = 6 \times 0 = 0\).
So the function \( f(x) = x^3\) has a inflexional point at x = 0.
In such case, the point is called saddle point

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Find the critical points

1. \(f(x)=8x^3+81x^2-42x-8\)

   Solution 👉 Click Here

   Given function is
   \(f(x)=8x^3+81x^2-42x-8\)
   Differentiating w. r. to x, we get
   \(f'(x)=24x^2+162x-42\)
   Solving the derivative for \(f'(x)=0\), we get
   \(f'(x)=0\)
   or\(24x^2+162x-42=0\)
   or\(x=0.25,x=-7\)
   So, the critical points are \(x=0.25,x=-7\).
2. \(f(x)=1+80x^3+5x^4-2x^5\)

   Solution 👉 Click Here

   Given function is
   \(f(x)=1+80x^3+5x^4-2x^5\)
   Differentiating w. r. to x, we get
   \(f'(x)=240x^2+20x^3-10x^4\)
   Solving the derivative for \(f'(x)=0\), we get
   \(f'(x)=0\)
   or\(240x^2+20x^3-10x^4=0\)
   or\(x^2=0, 24+2x-x^2=0\)
   or\(x=-4,x=0,x=6\)
   So, x=-4,x=0,x=6 are critical points.
   
3. \(f(x)=2x^3-7x^2-3x-2\)
4. \(f(x)=x^6-2x^5+8x^4\)
5. \(f(x)=4x^3-3x^2+9x+12\)
6. \(f(x)=\frac{x+4}{2x^2+x+8}\)

   Solution 👉 Click Here

   Given function is
   \(f(x)=\frac{x+4}{2x^2+x+8}\)
   Differentiating w. r. to x, we get
   \(f'(x)=\frac{-2(x^2+8x-2)}{(2x^2+x+8)^2}\)
   Solving the derivative for numirator=0, we get
   numirator=0
   or\(-2(x^2+8x-2)=0\)
   or\(x^2+8x-2\)
   or\(x=0.24,x=-8.24\)
   So, the critical points are \(x=0.24,x=-8.24\). Next, solving the derivative for denomirator=0, we get
   denomirator=0
   or\((2x^2+x+8)^2=0\)
   or\(2x^2+x+8=0\)
   orno real solution
   Hence, the critical points are \(x=0.24,x=-8.24\).
7. \(f(x)=\frac{1-x}{x^2+2x-15}\)

   Solution 👉 Click Here

   Given function is
   \(f(x)=\frac{1-x}{x^2+2x-15}\)
   Differentiating w. r. to x, we get
   \(f'(x)=\frac{x^2-2x+13}{(x^2+2x-15)^2}\)
   Solving the derivative for numirator=0, we get
   numirator=0
   or\(x^2-2x+13=0\)
   orno real solution
   Next, solving the derivative for denomirator=0, we get
   denomirator=0
   or\((x^2+2x-15)^2=0\)
   or\(x^2+2x-15=0\)
   or\(x=-5,x=3\)
   As we know, a critical point (derivative zero, DNE: derivative not exist) must exist as a point of original function. For this problem we found two values x=-5,x=3 where the derivative doesn’t exist, and the function also doesn’t exist, so neither of these are critical points.
   So, \(f(x)=\frac{1-x}{x^2+2x-15}\) have no critical points.
8. \(f(x)=\sqrt[5]{x^2-6x}\)

   Solution 👉 Click Here

   Given function is
   \(f(x)=\sqrt[5]{x^2-6x}\)
   Differentiating w. r. to x, we get
   \(f'(x)=\frac{2x-6}{5(x^2-6x)^{4/5}}\)
   Solving the derivative for numirator=0, we get
   numirator=0
   or\(2x-6=0\)
   or\(x=3\)
   Next, solving the derivative for denomirator=0, we get
   denomirator=0
   or\(5(x^2-6x)^{4/5}=0\)
   or\(x^2-6x=0\)
   or\(x=0,x=6\)
   As we know, a critical point (derivative zero, DNE: derivative not exist) must exist as a point of original function. For this problem we found two values x=0,x=3,x=6 exist in the function, so these x=0,x=3,x=6 are critical points.
   

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Find the inflexional point

1. Determine the concavity of \(f(x) = x^ 3 − 6 x^ 2 −12 x + 2\) and identify any points of inflection.

   Solution 👉 Click Here

   Given function is
   \(f(x) = x^ 3 − 6 x^ 2 −12 x + 2\)
   Differentiating w. r. to x, we get
   \(f'(x) = 3x^ 2 − 12 x −12\)
   Now, the second derivative is
   \(f''(x)=6x-12\)
   Solving the derivative for \(f''(x)=0\), we get
   \(f''(x)=0\)
   or\(6x-12=0\)
   or\(x=2\)
   The inflexional point is x=2.

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Find the concavity

1. Determine the concavity of \(f(x) = \sin x + \cos x\) on [0,2π] and identify any points of inflection

   Solution 👉 Click Here

   Given function is
   \(f(x) = \sin x + \cos x\)
   Differentiating w. r. to x, we get
   \(f'(x) = \cos x - \sin x\)
   Now, the second derivative is
   \(f''(x) = - \sin x - \cos x\)
   Solving the derivative for \(f''(x)=0\), we get
   \(f''(x)=0\)
   or\(- \sin x - \cos x=0\)
   or\(x=135^0, 315^0\)
   The inflexional point is \(x=\frac{3 \pi}{4}\) and \(x=\frac{7 \pi}{4}\) . Now

   interval	\((0,\frac{3 \pi}{4})\)	\((\frac{3 \pi}{4},\frac{7 \pi}{4})\)	\((\frac{7 \pi}{4},2\pi)\)
   \(f''(x)=- \sin x - \cos x\)	(-ve)	(+ve)	(-ve)
   result	down	up	down

   So, the curve is concave down during \((0,\frac{3 \pi}{4})\), concave up during \((\frac{3 \pi}{4},\frac{7 \pi}{4})\), and concave down during \((\frac{7 \pi}{4},2\pi)\).
2. Discuss the curve/function with respect to cancavity, point of inflexion, and local maxima, minima.
   1. \(y=x^4-4x^3\)
   2. \(y=x^{2/3}(6-x)^{1/3}\)
   3. \( f(x) = \frac{1}{6}x^4-2x^3+11x^2-18x\)
   4. \( f(x) = \frac{1}{12}x^4-2x^2\)
   5. \( f(x) = x^3-3x^2+1\)
   6. \( f(x) = \frac{2}{3}x^3-\frac{5}{2}x^2+2x\)

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Question for Practice

1. Find the stationary points of the function \( f(x) = 2x^3-3x^2-36x\).
2. Find the approximate values, to two decimal places, of the stationary points of the function \( f(x) = x^3-x^2-2x\).
3. Find the (a) intervals of increasing/decreasing (b) intervals of concave upward/downward (c) point of inflection
   1. \(f(x)=2x^3-3x^2-12x\)
   2. \( f(x) = \frac{4}{3}x^3 + 5x^2- 6x-2\)
   3. \(f(x)=2+3x-x^3\)
   4. \( f(x) = 3x^4 -2x^3- 9x^2 + 7\)
   5. \(f(x)=x^4-6x^2\)
   6. \(f(x)=200+8x^3+x^4\)
   7. \(f(x)=3x^5-5x^3+3\)
   8. \(f(x)=(x^2-1)^3\)
   9. \(f(x)=x \sqrt{x^2+1}\)
   10. \(f(x)=x-3x^{1/3}\)
4. Find the (a) vertical and horizental asymptotes (b) intervals of increasing/decreasing (c) intervals of concave upward/downward (d)point of inflection of following functions
   1. \(f(x)=\frac{1+x^2}{1-x^2}\)
   2. \(f(x)=\frac{x}{(x-1)^2}\)
   3. \(f(x)=\sqrt{x^2+1}-x\)

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Find the local Extreme

1. Find any turning points of the function \(f(x)=x^3-10x^2+25x+4\) and specify whether each turning point is a global maximum or minimum, or a local maximum or minimum.
   

   Solution 👉 Click Here

   Given function is
   \(f(x)=x^3-10x^2+25x+4\)
   Differentiating w. r. to x, we get
   \(f'(x)=3x^2-20x+25\)
   Solving the derivative for \(f'(x)=0\), we get
   \(f'(x)=0\)
   or\(3x^2-20x+25=0\)
   or\(x=5/3,x=5\)
   So, x=5/3,x=5 are critical points.
   Now, the second derivative is
   \(f''(x)=6x-20\)
   Now

   point	derivative	value	result
   at x=5/3	f''(x)=6x-20	-10 (-ve)	maxima
   at x=5	f''(x)=6x-20	10 (+ve)	minima

   So, x=5/3 is local maxima and x=5 is local minima.

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Question for Exam

1. Suppose you are given formula for a function f
   1. How do you determine where f is increasing/decreasing
   2. How do you determine where f is concave upward/downward
   3. How do you locate inflexional points
2. Find the critical numbers of \(f(x)=x^4(x-1)^3\).What does the second derivative tells you about the behavior of f at these points. What does the first derivative test tell you?
3. What is the use of first and second derivative in calculus? Explain geometrically. In a graph below, f, f ’ and f ’’ are shown, use first and second derivative to identify it with reasons “which is which”.

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Concavity of a function

In calculus, the second derivative of a function f is the derivative of f''. Roughly speaking, the second derivative measures instantaneous acceleration of the object, or the rate at which the velocity of the object is changing with respect to time.

On the graph of a function, the second derivative corresponds to the curvature or concavity of the graph. The graph of a function with a positive second derivative is upwardly concave, while the graph of a function with a negative second derivative curves in the opposite way.

1. The graph of \( y = f (x)\) is concave up if \( f'' > 0\)
2. The graph of \( y = f (x)\) is concave down if \( f'' < 0\)

The second derivative of a function f can be used to determine the concavity of the graph of f. A function whose second derivative is positive will be concave up (also referred to as convex), meaning that the tangent line will lie below the graph of the function. Similarly, a function whose second derivative is negative will be concave down (also simply called concave), and its tangent lines will lie above the graph of the function.

Second derivative test

The relation between the second derivative and the graph can be used to test whether a stationary point for a function (i.e., a point where \( f'(x)=0\) is a local maximum or a local minimum. Specifically,

1. If \( f''< 0\), then f has a local maximum at x.
2. If \( f''> 0\), then f has a local minimum at x.
3. If \( f''=0\), then the second derivative test says nothing about the point x, a possible inflection point.

Maxima and minima are known as the extrema of a function. For the function, under the entire range, the maximum value of the function is known as the global(absolute ) maxima and the minimum value is known as the global (absolute) minima.
There are other maxima and minima of a function, which are not the absolute maxima and minima of the function and are known as local maxima and local minima

Please note that:

1. The combination of maxima and minima is extrema.
2. global maxima is he highest point on the curve within the given range and minima is the lowest point on the curve.
3. local maxima can be less than local minima and vice-versa

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How to compute Maxima/Minima

1. Step 1: Find the turning point:
   Evaluate f ′(x) and solve f ′(x)=0.
   Say p=a,b,c,... are solutions (these are turning points).
2. Step 2: Evaluate f"(x) and find f"(p).
   If f "(p) < 0 then at x= p maxima, and f(p) is maximum value
   If f "(p)> 0 then at x=p minima, and f(p) is minimum value
   If f "(p)= 0 then test fails at x=p
   
3. Repeat this for all critical points p=a,b,c,...

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Exercise 1

1. Find the local minimum and maximum values of the function
   1. \(f(x) =x^3-6x^2-36x + 4\)

      Solution 👉 Click Here

      Given function is
      \( f(x) = x^3-6x^2-36x + 4\)
      Differentiating the function w.r. to x, we get
      \( f'(x) = 3x^2- 12x- 36\)
      Again, differentiating the function w.r. to x, we get
      \( f''(x) = 6x- 12\)
      For stationary point, we set
      \( f'(x) = 0\)
      or \( 3x^2- 12x- 36 = 0\)
      or \( x^2- 4x- 12 = 0\)
      or \( x = 6, x= - 2\)

      Now, the analytical table is given below.

      	x =-2	x = 6
      \( f''(x) = 6x- 12\)	f''(-2)=-24 (-ve)	f''(6)=24 (+ve)
      Result	Maximun value	Minimum value
      \( f(x) = x^3-6x^2-36x + 4\)	f(-2)=44	f(6)=-212

      
      The function f(x) has maximum value at x=-2, the maximum value is f(-2)=44.
      The function f(x) has minimum value at x=6, the minimumm value is f(6)=-212.
      
   2. \(f(x) =3x^4-4x^3-12x^2+5\)

      Solution 👉 Click Here

      Given function is
      \( f(x) = 3x^4-4x^3-12x^2+5\)
      Differentiating the function w.r. to x, we get
      \( f'(x) = 12x^3-12x^2-24x\)
      Again, differentiating the function w.r. to x, we get
      \( f''(x) = 36x^2-24x-24\)
      For stationary point, we set
      \( f'(x) = 0\)
      or \( 12x^3-12x^2-24x = 0\)
      or \( 12x(x^2-x-2) = 0\)
      or \( x = 0,x= - 1,x=2\)

      Now, the analytical table is given below.

      	x =-1	x = 0	x = 2
      \( f''(x) = 36x^2-24x-24\)	f''(-1)=36 (+ve)	f''(0)=-24 (-ve)	f''(2)=72 (+ve)
      Result	Minimum value	Maximum value	Minimum value
      \( f(x) = 3x^4-4x^3-12x^2+5\)	f(-1)=0	f(0)=5	f(2)=-27

      
      The function f(x) has minimum value at x=-1, the minimumm value is f(-1)=0
      The function f(x) has maximum value at x=0, the maximum value is f(0)=5
      The function f(x) has minimum value at x=2, the minimumm value is f(2)=-27
      
   3. \(f(x) =x +2 \sin x;0 \le x\le 2 \pi\)

      Solution 👉 Click Here

      Given function is
      \( f(x) = x +2 \sin x\)
      Differentiating the function w.r. to x, we get
      \( f'(x) = 1 +2 \cos x\)
      Again, differentiating the function w.r. to x, we get
      \( f''(x) = -2 \sin x\)
      For stationary point, we set
      \( f'(x) = 0\)
      or \( 1 +2 \cos x = 0\)
      or \(\cos x = -\frac{1}{2}\)
      or \( x = \frac{2 \pi}{3}, x= \frac{4 \pi}{3}\)

      Now, the analytical table is given below.

      	\( x = \frac{2 \pi}{3}\)	\( x = \frac{4 \pi}{3}\)
      \( f''(x) = -2 \sin x\)	\(f''(\frac{2 \pi}{3})=-\sqrt{3}\) (-ve)	\(f''(\frac{4 \pi}{3})=\sqrt{3}\) (+ve)
      Result	Maximun value	Minimum value
      \( f(x) = x +2 \sin x\)	\(f(\frac{2 \pi}{3})=3.83\)	\(f(\frac{4 \pi}{3})=2.46\)

      
      The function f(x) has maximum value at \( x = \frac{2 \pi}{3}\), the maximum value is \(f(\frac{2 \pi}{3})=3.83\)
      The function f(x) has minimum value at \( x = \frac{4 \pi}{3}\), the minimumm value is \(f(\frac{4 \pi}{3})=2.46\)
      
   4. \(f(x) =x^3-5x+3\)
   5. \(f(x) =\frac{x}{x^2+4}\)

      Solution 👉 Click Here

      Given function is
      \( f(x) =\frac{x}{x^2+4}\)
      Differentiating the function w.r. to x, we get
      \( f'(x) =\frac{4-x^2}{(x^2+4)^2}\)
      Again, differentiating the function w.r. to x, we get
      \( f''(x) = \frac{2x(x^2-12)}{(x^2+4)^3}\)
      For stationary point, we set
      \( f'(x) = 0\)
      or \( \frac{4-x^2}{(x^2+4)^2} = 0\)
      or \( x = -2, x= 2\)

      Now, the analytical table is given below.

      	\( x =-2\)	\( x =2\)
      \( f''(x) = \frac{2x(x^2-12)}{(x^2+4)^3}\)	\(f''(-2)=1/16\) (+ve)	\(f''(2)=-1/16\) (-ve)
      Result	Minimum value	Maximum value
      \( f(x) = \frac{x}{x^2+4}\)	\(f(-2)=-1/4\)	\(f(2)=1/4\)

      
      The function f(x) has maximum value at \( x = 2\), the maximum value is \(f(2)=1/4\)
      The function f(x) has minimum value at \( x =-2\), the minimumm value is \(f(-2)=-1.4\)
      
   6. \(f(x) =x+\sqrt{1-x}\)

   Solution 👉 Click Here

   Given function is
   \( f(x) =x+\sqrt{1-x}\)
   Differentiating the function w.r. to x, we get
   \( f'(x) =1-\frac{1}{2\sqrt{1-x}}\)
   Again, differentiating the function w.r. to x, we get
   \( f''(x) =-\frac{1}{4(1-x)^{3/2}}\)
   For stationary point, we set
   \( f'(x) = 0\)
   or \( 1-\frac{1}{2\sqrt{1-x}} = 0\)
   or \( x =3/4\)
   Now, the analytical table is given below.
   	\( x =3/4\)
   \( f''(x) =-\frac{1}{4(1-x)^{3/2}}\)	\(f''(3/4)=-2\) (-ve)
   Result	Maximum value
   \( f(x) = x+\sqrt{1-x}\)	\(f(3/4)=1.25\)
   
   The function f(x) has maximum value at \( x = 3/4\), the maximum value is \(f(3/4)=1.25\)
   
2. Show that \(f(x)= x^3-3x + 4\) is maximum when \( x =-1\) , minimum when \( x = 1\) , neither when \( x = 0\) .

   Solution 👉 Click Here

   Given function is
   \( f(x) = x^3-3x + 4\)
   Differentiating the function w.r. to x, we get
   \( f'(x) = 3x^2-3\)
   Again, differentiating the function w.r. to x, we get
   \( f''(x) =6x\)
   For stationary point, we set
   \( f'(x) = 0\)
   or \(3x^2-3 = 0\)
   or \( x^2 = 2\)
   or \( x = -1, x= 1\)

   Now, the analytical table is given below.

   	x =-1	x = 1
   \( f''(x) = 6x\)	f''(-1)=-6 (-ve)	f''(1)=6 (+ve)
   Result	Maximun value	Minimum value
   \( f(x) = x^3-3x + 4\)	f(-1)=6	f(1)=2

   
   The function f(x) has maximum value at x=-1, the maximum value is f(-1)=6.
   The function f(x) has minimum value at x=1, the minimumm value is f(1)=2.
   
3. Show that the maximum value of a function \( f (x) = x + \frac{100}{x}-25\) is less than its minimum value.

   Solution 👉 Click Here

   Given function is
   \( f(x) = x + \frac{100}{x}-25\)
   Differentiating the function w.r. to x, we get
   \( f'(x) = 1-\frac{100}{x^2}\)
   Again, differentiating the function w.r. to x, we get
   \( f''(x) =\frac{200}{x^3}\)
   For stationary point, we set
   \( f'(x) = 0\)
   or \(1-\frac{100}{x^2} = 0\)
   or \( x^2 = 100\)
   or \( x = -10, x= 10\)

   Now, the analytical table is given below.

   	x =-10	x = 10
   \( f''(x) = \frac{200}{x^3}\)	f''(-10)=-0.2 (-ve)	f''(10)=0.2 (+ve)
   Result	Maximun value	Minimum value
   \( f(x) = x + \frac{100}{x}-25\)	f(-10)=-45	f(10)=-5

   
   The function f(x) has maximum value at x=-10, the maximum value is f(-10)=-45.
   The function f(x) has minimum value at x=10, the minimumm value is f(10)=-5.
   
4. Show that the maximum value of a function \( f (x) = x^{5/3}-5x^{2/3}\) at x=0 is 0. Also find its minumum value at x=2.

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Test your understanding

🎯 Maxima and Minima Quiz

⏳ Loading 5 random questions...

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Exercise 2

Find the maximum and minimum values of following functions

1. \(f(x)=2x^3-3x^2-12x\)
2. \( f(x) = \frac{4}{3}x^3 + 5x^2- 6x-2\)
3. \(f(x)=2+3x-x^3\)
4. \(f(x)=200+8x^3+x^4\)
5. \(f(x)=x^4-6x^2\)
6. \(f(x)=3x^5-5x^3+3\)
7. \(f(x)=(x^2-1)^3\)
8. \(f(x)=x \sqrt{x^2+1}\)
9. \(f(x)=x-3x^{1/3}\)
10. \(f(x)=\frac{1+x^2}{1-x^2}\)
11. \(f(x)=\frac{x}{(x-1)^2}\)
12. \(f(x)=\sqrt{x^2+1}-x\)

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Exercise 3

1. Find the two numbers whose sum is 16 and the sum of whose squares is minimum.

   Solution 👉 Click Here

   Let \( x\) and \( y\) be two numbers such that
   \( x + y = 16\)
   If \( S\) be the sum of the squares of the two numbers, then
   \( S = x^2 + y^2\)
   or \( S = x^2 + (16- x)^2\) \( (y = 16- x)\) भएकोले
   Differentiating the function S w.r. to x, we get
   \( \frac{dS}{dx} = 4x- 32.\)
   Again, differentiating the function S' w.r. to x, we get
   \( \frac{d^2S}{dx^2} = 4\)
   For stationary point, set
   \( \frac{dS}{dx} = 0 \)
   or \( 4x- 32 = 0\)
   or \( x = 8 \)
   Since \( \frac{d^2S}{dx^2 } = 4 > 0\) , so \( S\) is minimum when \( x = 8\)
   Now, the corresponding value of \( y\) is
   \( y = 16- 8 = 8\)
   Hence the required numbers are \( 8,8\)
2. Find the dimensions of a rectangle with perimeter 100 metres so that the area of the rectangle is a maximum.
3. A rectangular box with a square base and no top is to have a volume of 108 cubic inches. Find the dimensions for the box that require the least amount of material.
4. A farmer has 24 m of fencing and wants to fence off a rectangular field that borders a straight river. He needs no fence along the river. What are the dimensions of the field that has the largest area?
5. A farmer with 80m of fencing wants to enclose a rectangular area and then divide it into four pens with fencing parallel to one side of the rectangle. What is the largest possible total area of the four pens?
6. A rectangular storage container With an open top is to have a volume of \(36 m^3\). The length of its base is twice the width. Material for the base costs Rs2 per square meter. Material for the sides costs Rs16 per square meter. Find the cost of materials for the cheapest such container.
7. Find the area of the largest rectangle that can be inscribed in a semicircle of radius r.
8. A cylindrical can is to be made to hold 1 L of oil. Find the dimensions that will minimize the cost of the metal to manufacture the can.
9. A right circular cylinder is inscribed in a right circular cone so that the center lines of the cylinder and the cone coincide. The cone has 8 cm and radius 6 cm. Find the maximum volume possible for the inscribed cylinder.
10. Suppose a company’s daily profit \(f(q)=100q-0.1q^2\) is the quantity of calculators sold per day. How many calculators should the company sell to maximise profit and what is the maximum daily profit?

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Derivative Ex 16.3 [BCB page 434]

1. Find, from the first principle, the derivative of:
   1. \( \log (ax+b)\)
   2. \( \log_5 x\)
   3. \( \log \frac{x}{10} \)
   4. \( e^{ax+b} \)
   5. \( e^{\frac{x}{3}} \)
2. Find the derivative of:
   1. \( \log (/sin x)\)
   2. \( \log (x+/tan x)\)
   3. \( \log (1+e^{5x})\)
   4. \( \log (\log x)\)
   5. \( \log (\sec x)\)
   6. \( \log (1+\sin^2 x)\)
   7. \( \ln (e^{ax}+e^{-ax})\)
   8. \( \log (\sqrt{a^2+x^2}+b)\)
   9. \( \log (\sqrt{a+x}+(\sqrt{a-x})\)
   10. \( \ln |x-4|\)
3. Find the differential coefficient of:
   1. \( e^{\sin x} \)
   2. \( e^{\sqrt{\cos x}} \)
   3. \( e^{1+\log x} \)
   4. \( e^{\sin (\log x)} \)
   5. \( \tan (\log x) \)
   6. \( \sin (1+e^{ax}) \)
   7. \( \cos (\log \sec x) \)
   8. \( \sec (\log \tan x) \)
   9. \( \sin \log \sin e^{x^2} \)
4. Differentiate the following with respect to x:
   1. \( x^2 \log (1+x)\)
   2. \( x^5 e^{ax}\)
   3. \( \sin ax \log x\)
   4. \( e^{ax} \cos bx\)
   5. \( (\tan x+x^2)\log x\)
   6. \( \sin x+\cos x) e^{ax} \)
5. Calculate the derivative of:
   1. \( \frac{\log x }{\sin x} \)
   2. \( \frac{\log (ax+b) }{e^{px}} \)
   3. \( \frac{e^{ax} }{\cos bx} \)
   4. \( \frac{\sin ax }{1+\log x} \)
   5. \( \frac{\log x }{a^2+x^2 } \)
6. Find \(\frac{dy}{dx}\) when
   1. \( xy=\log (x^2+y^2) \)
   2. \( x^2+y^2=\log (x+y)\)
   3. \( e^{xy}=xy \)
   4. \( x=e^{\cos 2t}, y=e^{\sin 2t} \)
   5. \( x=\cos (\log t), y=\log(\cos t) \)
   6. \( x=\log t+\sin t, y=e^t+\cos t \)

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Practical Question

A function \(y=f(x)\) is considered, say \(y=\), now answer to the following questions.

1. Draw the graph of the function \(y=f(x)\)
2. Given the point on the curve \(y=f(x)\), draw a tangent at the point making an angle \(\theta\) with the positive direction of x-axis
3. Find \(\tan \theta\)
4. Find the point on the curve where the tangent is parallel to the x-axis, if possible
5. Find the point on the curve where the tangent is perpendicular to the x-axis, if possible

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Additional Question

1. Define differential cofficient of a function at a given point. Find from first principles, the derivative of \( x |\sqrt{x}\) with respect to x
2. Find the derivative of
   1. \( x^{\sin x} \)

      Solution 👉 Click Here

      Let
      \( y =x^{\sin x}\)
      
      Taking log on both sides, we get
      
      \( \log y =\log x^{\sin x}\)
      
      or\( \log y =\sin x \log x \)\(\log a^b=b \log a\) भएकोले
      
      Differentiating both side w.r.to x, we get
      
      \( \frac{d}{dx}\log y =\frac{d}{dx} (\sin x \log x )\)
      
      or\( \frac{1}{y} \frac{dy}{dx} =\sin x \left (\frac{d}{dx}\log x \right ) + \left (\frac{d}{dx} \sin x \right ) \log x \)
      
      or\( \frac{1}{y} \frac{dy}{dx} =\sin x \left (\frac{1}{x} \right ) + \left ( \cos x \right ) \log x \)
      
      or\(\frac{dy}{dx} = y \left (\frac{\sin x }{x}+ \cos x \log x \right ) \)
      
      or\(\frac{dy}{dx} = x^{\sin x} \left (\frac{\sin x }{x}+ \cos x \log x \right ) \)
      

   2. \( (\sin x)^x \)

      Solution 👉 Click Here

      Let
      \( y =(\sin x)^x\)
      
      Taking log on both sides, we get
      
      \( \log y =\log (\sin x)^x\)
      
      or\( \log y =x \log (\sin x) \)\(\log a^b=b \log a\) भएकोले
      
      Differentiating both side w.r.to x, we get
      
      \( \frac{d}{dx}\log y =\frac{d}{dx} [x \log (\sin x) ]\)
      
      or\( \frac{1}{y} \frac{dy}{dx} =x \left [\frac{d}{dx} \log (\sin x) \right ] + \left [\frac{d}{dx} x \right ]\log (\sin x)\)
      
      or\( \frac{1}{y} \frac{dy}{dx} =x \left [\frac{1}{\sin x} \cos x\right ] + \left [ 1 \right ] \log (\sin x) \)
      
      or\(\frac{dy}{dx} = y \left [x \cot x +\log (\sin x)\right ] \)
      
      or\(\frac{dy}{dx} = (\sin x)^x \left [x \cot x +\log (\sin x)\right ]\)
      

   3. \( (\sin x)^{\log x} \)

      Solution 👉 Click Here

      Let
      \( y =(\sin x)^{\log x}\)
      
      Taking log on both sides, we get
      
      \( \log y =\log (\sin x)^{\log x} \)
      
      or\( \log y =\log x \log (\sin x) \)\(\log a^b=b \log a\) भएकोले
      
      Differentiating both side w.r.to x, we get
      
      \( \frac{d}{dx}\log y =\frac{d}{dx} [\log x \log (\sin x) ]\)
      
      or\( \frac{1}{y} \frac{dy}{dx} =\log x \left [\frac{d}{dx} \log (\sin x) \right ] + \left [\frac{d}{dx} \log x \right ]\log (\sin x)\)
      
      or\( \frac{1}{y} \frac{dy}{dx} =\log x \left [\frac{1}{\sin x} \cos x\right ] + \left [\frac{1}{x} \right ] \log (\sin x) \)
      
      or\(\frac{dy}{dx} = y \left [\log x \cot x + \frac{1}{x}\log (\sin x)\right ] \)
      
      or\(\frac{dy}{dx} = (\sin x)^{\log x}\left [\log x \cot x + \frac{1}{x}\log (\sin x)\right ]\)
      

   4. \( e^{x^x} \)

      Solution 👉 Click Here

      Let
      \( y =e^{x^x}\)
      
      Taking log on both sides, we get
      
      \( \log y =\log e^{x^x} \)
      
      or\( \log y =x^x\log e\)\(\log a^b=b \log a\) भएकोले
      
      or\( \log y =x^x \)
      
      Taking log on both sides, we get
      
      \( \log \log y =\log x^x\)
      
      or\( \log \log y = x \log x \)\(\log a^b=b \log a\) भएकोले
      
      Differentiating both side w.r.to x, we get
      
      \( \frac{d}{dx} \log \log y =\frac{d}{dx} [x \log x ]\)
      
      or\( \frac{1}{\log y} \frac{1}{y} \frac{dy}{dx} = x \left [\frac{d}{dx} \log x \right ] + \left [\frac{d}{dx} x \right ]\log x\)
      
      or\( \frac{1}{y \log y} \frac{dy}{dx} =(1+\log x) \)
      
      or\(\frac{dy}{dx} = y \log y (1+\log x) \)
      
      or\(\frac{dy}{dx} = e^{x^x} x^x (1+\log x)\)\(\log y=x^x\) भएकोले
      
      

   5. \( x^{e^x} \)
   6. \( (\log x)^{\tan x} \)
   7. \( (\tan x)^{\log x} \)
   8. \( x^{\sec x} \)
   9. \( (\sin x)^{\cos x} \)
3. Find \(\frac{dy}{dx}\) when
   1. \( y=x^y \)
   2. \( x^y.y^x=1\)
   3. \( x^m.y^n=(x+y)^{m+n}\)
   4. \( e^{\sin x}+e^{\sin y}=1\)
   5. \(x^y=y^x\)
   6. \(x^{\sin x}=y^{\sin y}\)