Derivative

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Introduction to Derivative

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 ${\displaystyle f^{\prime }(x)}$ or $y^{\prime }$.
   Joseph-Louis Lagrange ले derivative लाई ${\displaystyle f^{\prime }(x)}$ or $y^{\prime }$ prime notation बाट जनाएका थिए।
3. Newton ले ले derivative लाई 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.

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Introduction to Slope

सामान्यतयाः

gradient =$\frac{\text{change in y}}{\text{change in x}}$

भन्ने बुझिन्छ। जसलाई तलको चित्रबाट हेरौ।

0,0

–o+←↓↑→

2.0

A

2.0

Fig.1

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 हरु को मान बढ्छ, घट्छ वा उस्तै रहन्छ भन्नेमा निर्भर गर्दछ।
खास गरी, बायाँबाट दायाँ जादा रेखाको स्थिति तल गएमा ऋणात्मक ग्रेडियन्ट (slope), तेर्सो भएमा शून्य ग्रेडियन्ट (slope) हुन्छ, र माथि गएमा धनात्मक ग्रेडियन्ट (slope) हुन्छ ।
यदि रन (Run) को मान शून्य छ भने रेखाको ग्रेडियन्ट (slope) अपरिभाषित हुन्छ, किनकि शून्यले भाग गर्ने कुरा सम्भव छैन। त्यसैले ठाडो रेखाहरूको ग्रेडिएन्टहरू अपरिभाषित हुन्छन्।
गणितिय रुपमा, दुईवटा बिन्दुहरू $(x_1,y_1)$ र $(x_2,y_2)$ छन भने ति बिन्दुहरुले बनाउने सिधा रेखाको ग्रेडियन्ट (slope) तलको चित्रमा चित्रण गरे जस्तै (चित्र 2) हुन्छ।

0,0

–o+←↓↑→

$(x_1,y_1)$

$(x_2,y_2)$

Run=$x_2-x_1$

Rise=$y_2-y_1$

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, $(x_1,y_1)$, and which you take to be the second point, $(x_2,y_2)$, as we get the same result either way.

Example 1

For example, using the formula to calculate the gradient of the line through the points (1, 8) and (5, 2), which is shown in Figure below [Figure 3]

0,0

(1,8)

(5,2)

Fig.3

It gives either
gradient =$\frac{y_2-y_1}{x_2-x_1}=\frac{2-8}{5-1}=-\frac{3}{2}$
gradient =$\frac{y_2-y_1}{x_2-x_1}=\frac{8-2}{1-5}=-\frac{3}{2}$

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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. It's essentially the slope of the function at that point. When we say the derivative represents the rate of instantaneous rate of change, we mean that it gives us the rate at which the function is changing at a specific instant or point.

0,0

–o+←↓↑→

P

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=$\frac{f(a+\delta )-f(a-\delta )}{2\delta }$ For example, when we need the change in function $f(x)=x^2$ at an instant point , say x=2, then the instantaneous rate of change at x=2 is given by
change=$\frac{f(2+0.1)-f(2-0.1)}{2\times 0.1}=\frac{f(2.1)-f(1.9)}{0.2}=\frac{{2.1}^2-{1.9}^2}{0.2}$

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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.

0,0

–o+←↓↑→

P(x,f(x))

Q(x+h,f(x+h))

Fig.4

Let y = 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 given by
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 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 $P={\displaystyle \underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}}$
or slope of tangent at P = f'(x)
This new function f'(x) is called the derivative (or derived function) of the function f(x), and is denoted by f′(x) (which is read as‘f prime’ or ‘f dash’ or ‘f dashed’).
The process of finding the derivative of a given function f(x) is called differentiation, and when we carry out this process, we say that we’re differentiating the function f(x).
For example, a function is $f(x)=x^2$, then its derivative is f'(x) = 2x.

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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: $f(x+\mathrm{\Delta }x)$
step3: $f(x+\mathrm{\Delta }x)-f(x)$
step4: $\frac{f(x+\mathrm{\Delta }x)-f(x)}{\mathrm{\Delta }x}$
step5: Simplify $\frac{f(x+\mathrm{\Delta }x)-f(x)}{\mathrm{\Delta }x}$
step6: ${\displaystyle \underset{\mathrm{\Delta }x\to 0}{lim}\frac{f(x+\mathrm{\Delta }x)-f(x)}{\mathrm{\Delta }x}}$

Example 2

Use the definition of the derivative to compute the derivative of $f(x)=x^n$
The solution is
$f^{\prime }(x)={\displaystyle \underset{\mathrm{\Delta }x\to 0}{lim}\frac{f(x+\mathrm{\Delta }x)-f(x)}{\mathrm{\Delta }x}}$
or $f^{\prime }(x)={\displaystyle \underset{\mathrm{\Delta }x\to 0}{lim}\frac{{(x+\mathrm{\Delta }x)}^n-x^n}{\mathrm{\Delta }x}}$
or $f^{\prime }(x)={\displaystyle \underset{\mathrm{\Delta }x\to 0}{lim}\frac{(x^n+n{x}^{n-1}\mathrm{\Delta }x+...+x\mathrm{\Delta }{x}^{n-1}+\mathrm{\Delta }{x}^n)-x^n}{\mathrm{\Delta }x}}$
or $f^{\prime }(x)={\displaystyle \underset{\mathrm{\Delta }x\to 0}{lim}\frac{n{x}^{n-1}\mathrm{\Delta }x+...+x\mathrm{\Delta }{x}^{n-1}+\mathrm{\Delta }{x}^n}{\mathrm{\Delta }x}}$
or $f^{\prime }(x)={\displaystyle \underset{\mathrm{\Delta }x\to 0}{lim}\frac{\mathrm{\Delta }x(n{x}^{n-1}+...+x\mathrm{\Delta }{x}^{n-2}+\mathrm{\Delta }{x}^{n-1})}{\mathrm{\Delta }x}}$
or $f^{\prime }(x)={\displaystyle \underset{\mathrm{\Delta }x\to 0}{lim}(n{x}^{n-1}+...+x\mathrm{\Delta }{x}^{n-2}+\mathrm{\Delta }{x}^{n-1})}$
or $f^{\prime }(x)=n{x}^{n-1}$

Example 3

Find the derivative of $f(x)=\sin x$ by first principle.
By definition, the derivative of $f(x)$ is
$f^{\prime }(x)={\displaystyle \underset{\mathrm{\Delta }x\to 0}{lim}\frac{f(x+\mathrm{\Delta }x)-f(x)}{\mathrm{\Delta }x}}$
or $f^{\prime }(x)={\displaystyle \underset{\mathrm{\Delta }x\to 0}{lim}\frac{\sin (x+\mathrm{\Delta }x)-\sin x}{\mathrm{\Delta }x}}$
or $f^{\prime }(x)={\displaystyle \underset{\mathrm{\Delta }x\to 0}{lim}\frac{2\cos (\frac{x+\mathrm{\Delta }x+x}{2}).\sin (\frac{x+\mathrm{\Delta }x-x}{2})}{\mathrm{\Delta }x}}$
or $f^{\prime }(x)={\displaystyle \underset{\mathrm{\Delta }x\to 0}{lim}\frac{2\cos (x+\frac{\mathrm{\Delta }x}{2}).\sin (\frac{\mathrm{\Delta }x}{2})}{\mathrm{\Delta }x}}$
or $f^{\prime }(x)={\displaystyle \underset{\mathrm{\Delta }x\to 0}{lim}\cos (x+\frac{\mathrm{\Delta }x}{2}).\frac{\sin (\frac{\mathrm{\Delta }x}{2})}{\frac{\mathrm{\Delta }x}{2}}}$
or $f^{\prime }(x)={\displaystyle \underset{\mathrm{\Delta }x\to 0}{lim}\cos (x+\frac{\mathrm{\Delta }x}{2}).{\displaystyle \underset{\mathrm{\Delta }x\to 0}{lim}\frac{\sin (\frac{\mathrm{\Delta }x}{2})}{\frac{\mathrm{\Delta }x}{2}}}}$
or $f^{\prime }(x)=\cos (x+\frac{0}{2}).1$
or $f^{\prime }(x)=\cos x$

Example 4

Find the derivative of $f(x)=e^x$ by first principle.
By definition, the derivative of $f(x)$ is
$f^{\prime }(x)={\displaystyle \underset{\mathrm{\Delta }x\to 0}{lim}\frac{f(x+\mathrm{\Delta }x)-f(x)}{\mathrm{\Delta }x}}$
or $f^{\prime }(x)={\displaystyle \underset{\mathrm{\Delta }x\to 0}{lim}\frac{e^{x+\mathrm{\Delta }x}-e^x}{\mathrm{\Delta }x}}$
or $f^{\prime }(x)={\displaystyle \underset{\mathrm{\Delta }x\to 0}{lim}\frac{e^x(e^{\mathrm{\Delta }x}-1)}{\mathrm{\Delta }x}}$
or $f^{\prime }(x)=e^x{\displaystyle \underset{\mathrm{\Delta }x\to 0}{lim}\frac{e^{\mathrm{\Delta }x}-1}{\mathrm{\Delta }x}}$
or $f^{\prime }(x)=e^x$

Example 5

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

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 $\frac{f(x+h)-f(x)}{h}=\frac{h+\frac{1}{x+h}-\frac{1}{x}}{h}=\frac{x^2+hx-1}{x(h+x)}$
step6: ${\displaystyle \underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}=1-\frac{1}{x^2}}$

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Application of Derivative

1. Equation of tangent line

   Tangent is limiting position of secant
   0,0

   –o+←↓↑→

   P

   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.

   0,0 –o+←↓↑→ $$A=(x_1,y_1)$$	0,0 –o+←↓↑→ A B
   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

2. 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 $f^{\prime }(x)$ 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 $(a,b)$ .

   0,0

   –o+←↓↑→

   $$(x_1,f(x_1))$$

   $$(x_2,f(x_2))$$

   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 $f^{\prime }(x)$ 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 $(a,b)$ .

   0,0

   –o+←↓↑→

   $$(x_1,f(x_1))$$

   $$(x_2,f(x_2))$$

   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 $x$.

   0,0

   –o+←↓↑→

   f(x)

   f'(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).

   0,0

   –o+←↓↑→

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

   	(-∞,-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 $f^{\prime }(x)=6(x+2)(x-3)$ 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 $f^{\prime }(x)=6(x+2)(x-3)$ 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 $f^{\prime }(x)=6(x+2)(x-3)$ is also positive.
      Therefore, by the increasing/decreasing criterion, the function f is increasing on the interval (3,∞).
   First Derivative Test
   0,0

   0.26

   f'(x)= 0.26

3. Nature of points

   Critical Point

   Critical points are places where the derivative of a function is either zero or undefined. that is, either the graph of function has a horizontal tangent line or function is not differentiable at the point.
   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
   0,0

   A

   B

   C

   D

   Stationary Point

   Stationary points are places where the derivative of a function is zero (only zero), that is, the graph of function has a horizontal tangent line at the 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
   0,0

   A

   B

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

   Turning Point

   A stationary point where the derivative of the function changes sign (from positive to negative, or vice versa) at that point. Thus, a stationary point in which a function has a local maximum or local minimum is called a turning point.
   There are two types of turning points.

   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 not that A and B is not Turning point] in which

   1. f'(x)=0 at A, f''(x)=0 at A
   2. f'(x) is NOt defined at B B
   3. f'(x)=0 at C, local minima is defined at C
   0,0

   A

   B

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

   Straight line

   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 a horizontal point of inflection. It is a straight line.

   0,0

   Point of Inflection

   If the second derivative of a function changes sign, the graph of the function will switch from concave down to concave up, or vice versa. A point where this occurs 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.

   0,0

   f''(x)= 0.73

   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.

   Saddle point (Horizontal point of inflection)

   A saddle point or minimax point is a point on the graph of 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 $f^{\prime }(x)=3x^2$, so the gradient of the graph when x = 0 is
   $f^{\prime }(0)=3\times 0^2=0$.
   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 is $f^{″}(x)=6x$, so the second derivative of the graph 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

   0,0

4. 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.

   0,0

   f''(x)= 0.73

   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 $f^{\prime }(x)=0$ 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.

5. Shape of graph

6. Maxima and Minima of a function

7. Approximation

8. Rolls' Theorem

9. Mean Value Theorem

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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. $\log (/sinx)$
   2. $\log (x+/tanx)$
   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$\frac{1}{y}\frac{dy}{dx}=\sin x(\frac{d}{dx}\log x)+(\frac{d}{dx}\sin x)\log x$
      
      or$\frac{1}{y}\frac{dy}{dx}=\sin x\left(\frac{1}{x}\right)+(\cos x)\log x$
      
      or$\frac{dy}{dx}=y(\frac{\sin x}{x}+\cos x\log x)$
      
      or$\frac{dy}{dx}=x^{\sin x}(\frac{\sin x}{x}+\cos x\log x)$
      

   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$\frac{1}{y}\frac{dy}{dx}=x[\frac{d}{dx}\log (\sin x)]+\left[\frac{d}{dx}x\right]\log (\sin x)$
      
      or$\frac{1}{y}\frac{dy}{dx}=x[\frac{1}{\sin x}\cos x]+\left[1\right]\log (\sin x)$
      
      or$\frac{dy}{dx}=y[x\cot x+\log (\sin x)]$
      
      or$\frac{dy}{dx}=(\sin x{)}^x[x\cot x+\log (\sin x)]$
      

   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$\frac{1}{y}\frac{dy}{dx}=\log x[\frac{d}{dx}\log (\sin x)]+[\frac{d}{dx}\log x]\log (\sin x)$
      
      or$\frac{1}{y}\frac{dy}{dx}=\log x[\frac{1}{\sin x}\cos x]+\left[\frac{1}{x}\right]\log (\sin x)$
      
      or$\frac{dy}{dx}=y[\log x\cot x+\frac{1}{x}\log (\sin x)]$
      
      or$\frac{dy}{dx}=(\sin x{)}^{\log x}[\log x\cot x+\frac{1}{x}\log (\sin x)]$
      

   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$\frac{1}{\log y}\frac{1}{y}\frac{dy}{dx}=x[\frac{d}{dx}\log x]+\left[\frac{d}{dx}x\right]\log x$
      
      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. $x^{\sin x}=y^{\sin y}$