Differentiation (Part-38) | S N De

Differentiation (Part-35) | S N De

 

Here are five key points about differentiation in mathematics:

1. Rate of Change: Differentiation allows us to calculate the rate at which a function changes with respect to its independent variable. By finding the derivative of a function, we can determine the instantaneous rate of change of the function at any given point. This is particularly useful for studying phenomena that involve varying quantities over time or space.

2. Tangent Lines: The derivative of a function gives us the slope of the tangent line to the curve at a specific point. This tangent line represents the best linear approximation to the curve near that point. Tangent lines help us understand the local behavior and direction of a curve.

3. Higher Order Derivatives: Differentiation can be applied repeatedly to find higher order derivatives. The second derivative represents the rate of change of the slope (or curvature) of the original function, while the third derivative represents the rate of change of the curvature itself. Higher order derivatives provide information about the concavity and shape of the curve.

4. Optimization: Differentiation plays a crucial role in optimization problems. By finding the critical points of a function (where the derivative is zero or undefined), we can determine maximum and minimum values. This helps in maximizing profits, minimizing costs, optimizing resource allocation, and solving various real-world problems.

5. Differentiation Rules: There are several rules and formulas that facilitate the process of finding derivatives. These include the power rule, product rule, quotient rule, and chain rule, among others. These rules provide systematic methods to differentiate different types of functions, making the process more efficient and manageable.

Overall, differentiation is a fundamental concept in calculus that allows us to understand the rates of change, approximate curves locally, solve optimization problems, and analyze the behavior and properties of functions. It provides powerful tools for studying a wide range of mathematical and real-world phenomena.

Now, we are going to solve few problems in S N Dey Mathematics (Class -12 )

$~2(vi)~~f(x)=\cos^2x~~$ in $~~-\frac{\pi}{4} \leq x \leq  \frac{\pi}{4} .$

Sol. Clearly $~f(x)~$ is a trigonometric function $~\cos^2x~$ is continuous in $~[-\pi/4,\pi/4]~$ and differentiable in $~(-\pi/4,\pi/4).$ 

Also, $~~f(-\pi/4)=\cos^2(\pi/4) =(\frac{1}{\sqrt 2})^2=\frac 12, \\ ~~f(\pi/4)=\cos^2(\pi/4)=(\frac{1}{\sqrt 2})^2=\frac 12 \\ \therefore ~~f(-\pi/4)=f(\pi/4).$

So, the given function satisfies all conditions of Rolle's theorem.

$~\therefore~$  there exists at least one value of $~x~$ say $~x=c \in (-\pi/4,\pi/4)~$ such that $~~f'(c)=0 \rightarrow(1)$

Now, $~~f'(x)=\frac{d}{dx}(\cos^2x)=2\cos x~\cdot(-\sin x) \\ \therefore f'(x)=-\sin2x \\ \text{so,}~~f'(c)=-\sin 2c \\ \text{or,}~~0=\sin 2c~~[\text{By (1)}]\\ \therefore \sin 2c=0=\sin(n\pi)~~[~n \in \mathbb Z] \\ \text{or,}~ 2c=n\pi \\ \text{or,}~ c=\frac{n\pi}{2} \\~\therefore~\text{for }~n=0, ~~c=0\in (-\pi/4,\pi/4)$

Hence, there exists one value of $~c=0~$ in the given interval for which $~~f'(c)=0~~$ and so, Rolle's theorem is verified.

Note : $~~\mathbb{Z} \rightarrow~$ the set of integers.

$~(vii)~f(x)=x^2-x-12~~$ in $~~[-3,4]$

Sol. Sol.  Since a polynomial function is continuous and differentiable in any finite interval, clearly, $~f(x)~$  being a polynomial function, is continuous in $~~-3 \leq x \leq 4~,$ differentiable in $~-3<x<4~.$

Also, $~f(-3)=(-3)^2-(-3)-12\\~~~~~~~~~~~=9+3-12=0\\~~f(4)=(4)^2-(4)-12\\~~~~~~~~~~~=16-4-12=0 \\ \therefore~~f(-3)=f(4)$

So, the given function satisfies all conditions of Rolle's theorem. $~\therefore~$  there exists at least one value of $~x~$ say $~x=c \in (-3,4)~$ such that $~~f'(c)=0 \rightarrow(1)$ 

Now, $~~f'(x)\\=\frac{d}{dx}(x^2-x-12)\\=2x-1 \rightarrow(2)$

So, by $~(1)~$ and $~(2)~$ we get,

$~~~f'(c)=0\\ \text{or,}~~2c-1=0 \\ \text{or,}~~c=\frac 12 \in(-3,4)$

Hence, there exists one value of $~c=\frac 12~$ in the given interval for which $~~f'(c)=0~~$ and so, Rolle's theorem is verified.

$~(viii)~f(x)=x^2-5x+6~~$ in $~~2 \leq x \leq 3$

Sol.  Since a polynomial function is continuous and differentiable in any finite interval, clearly, $~f(x)~$  being a polynomial function, is continuous in $~~2 \leq x \leq 3~,$ differentiable in $~2<x<3~.$

Also, $~f(2)=(2)^2-5 \times 2+6\\~~~~~~~~=4-10+6=0,\\~~f(3)=(3)^2-5 \times 3+6\\~~~~~~~~~=9-15+6=0 \\ \therefore~~f(2)=f(3)$

So, the given function satisfies all conditions of Rolle's theorem. $~\therefore~$  there exists at least one value of $~x~$ say $~x=c \in (2,3)~$ such that $~~f'(c)=0 \rightarrow(1)$ 

Now, $~~f'(x)\\=\frac{d}{dx}(x^2-5x+6)\\=2x-5 \rightarrow(2)$

So, by $~(1)~$ and $~(2)~$ we get,

$~~~f'(c)=0\\ \text{or,}~~2c-5=0 \\ \text{or,}~~c=\frac 52 \in(2,3)$

Hence, there exists one value of $~c=\frac 52~$ in the given interval for which $~~f'(c)=0~~$ and so, Rolle's theorem is verified.

$~(ix)~~f(x)=\cos x~~$ in $~\left[-\frac{\pi}{2},\frac{\pi}{2}\right].$

Sol. Clearly $~f(x)~$ is a trigonometric function $~\cos x~~$ is continuous in $~\left[-\frac{\pi}{2},\frac{\pi}{2}\right]~$ and differentiable in $~(-\frac{\pi}{2},\frac{\pi}{2}).$ 

Also, $~~f(-\pi/2)=\cos(-\pi/2) =0,\\~~f(\pi/2)=\cos(\pi/2)=0\\ \therefore~f(-\pi/2)=f(\pi/2).$

So, the given function satisfies all conditions of Rolle's theorem.

$~\therefore~$  there exists at least one value of $~x~$ say $~x=c \in (-\pi/2,\pi/2)~$ such that $~~f'(c)=0 \rightarrow(1)$

Now, $~~f'(x)=\frac{d}{dx}(\cos x)\\ \therefore f'(x)=-\sin x \\ \text{so,}~~f'(c)=-\sin c \\ \text{or,}~~0=-\sin c~~[\text{By (1)}]\\ \therefore~ \sin c=0=\sin n\pi~~[\text{where}~n \in \mathbb{Z}] \\ \text{so for,}~n=0,\\~~ c=0 \times \pi=0  \in (-\pi/2,\pi/2)~$

Hence, there exists one value of $~c=0~$ in the given interval for which $~~f'(c)=0~~$ and so, Rolle's theorem is verified.

$~(x)~~f(x)=\sin 3x~~$ in $~0 \leq x \leq \pi.$

Sol. Clearly $~f(x)~$ is a trigonometric function $~\sin 3x~~$ is continuous in $~\left[0,\pi\right]~$ and differentiable in $~(0,\pi).$ 

Also, $~~f(0)=\sin 0=0,\\~~f(\pi)=\sin3\pi=0\\ \therefore~f(0)=f(\pi).$

So, the given function satisfies all conditions of Rolle's theorem.

$~\therefore~$  there exists at least one value of $~x~$ say $~x=c \in (0,\pi)~$ such that $~~f'(c)=0 \rightarrow(1)$

Now, $~~f'(x)=\frac{d}{dx}(\sin 3x)\\ \therefore f'(x)=3\cos 3x\\ \text{so,}~~f'(c)=3\cos 3c \\ \text{or,}~~0=3\cos 3c~~[\text{By (1)}]\\ \therefore~ \cos 3c=0\\ \therefore~ 3c=(2n+1)\pi/2~~[\text{where}~n \in \mathbb{Z}] \\ \text{so for,}~n=0,\\~~ 3c=\pi/2 \\ \text{or,}~~c=\frac{\pi}{6}  \in (0, \pi)~$

Hence, there exists one value of $~c=\frac{\pi}{6}~$ in the given interval for which $~~f'(c)=0~~$ and so, Rolle's theorem is verified.

$~(xi)~~f(x)=e^{-x}\sin x~~$ in $~[0,\pi].$

Sol. We know that the exponential function $~e^{-x}~$ and trigonometric function $~\sin x~$ are continuous for every $~~x \in \mathbb{R} ;$ hence the product function $~f(x)=e^{-x}\sin x~~$ is continuous in $~[0,\pi].$

Again, $~f'(x)=\frac{d}{dx}(e^{x}\sin x)=\frac{d}{dx}(e^{-x})\cdot \sin x+ \frac{d}{dx}(\sin x)\cdot e^{-x} \\ \therefore f'(x)=-e^{-x}\sin x+e^{-x} \cos x \\ \text{or,}~~f'(x)=e^{-x}(\cos x-\sin x)\rightarrow(1)$

Clearly, $~f'(x)~$ exists everywhere in $~(0,\pi).$

Further, $~f(0)=e^0\sin 0=0, \\~~f(\pi)=e^{-\pi}\sin \pi=0 \\ \therefore~~f(0)=f(\pi).$

So, the given function satisfies all conditions of Rolle's theorem.

$~\therefore~$  there exists at least one value of $~x~$ say $~x=c \in (0,\pi)~$ such that $~~f'(c)=0 \rightarrow(2)$

So, from $~(1),~(2)~$ we get,

$~~f'(c)=0 \\ \text{or,}~~e^{-c}(\cos c-\sin c)=0 \\ \text{or,}~~ \cos c-\sin c=0~~[\because~e^{-c} \neq 0] \\ \text{or,}~~ \cos c=\sin c \\ \text{or,}~~ \frac{\sin c}{\cos c}=1 \\ \text{or,}~~ \tan c=1=\tan(\pi/4) \\ \therefore~c=\frac{\pi}{4} \in (0,\pi).$

Hence, there exists one value of $~c=\frac{\pi}{4}~$ in the given interval for which $~~f'(c)=0~~$ and so, Rolle's theorem is verified.

$~(xii)~f(x)=x^2-4x+3~~$ in $~~[1,3]$

Sol.  Since a polynomial function is continuous and differentiable in any finite interval, clearly, $~f(x)~$  being a polynomial function, is continuous in $~~1 \leq x \leq 3~,$ differentiable in $~1<x<3~.$

Also, $~f(1)=1^2-4\times 1+3\\~~~~~~~~=1-4+3=0,\\~~f(3)=(3)^2-4 \times 3+3\\~~~~~~~~~=9-12+3=0 \\ \therefore~~f(1)=f(3)$

So, the given function satisfies all conditions of Rolle's theorem. $~\therefore~$  there exists at least one value of $~x~$ say $~x=c \in (1,3)~$ such that $~~f'(c)=0 \rightarrow(1)$ 

Now, $~~f'(x)\\=\frac{d}{dx}(x^2-4x+3)\\=2x-4 \rightarrow(2)$

So, by $~(1)~$ and $~(2)~$ we get,

$~~~f'(c)=0\\ \text{or,}~~2c-4=0 \\ \text{or,}~~c=\frac 42=2 \in(1,3)$

Hence, there exists one value of $~c=2~$ in the given interval for which $~~f'(c)=0~~$ and so, Rolle's theorem is verified.

Errorless 45 Previous Years IIT JEE Advanced (1978  2021) + JEE Main (2013  2022) PHYSICS Chapterwise & Topicwise Solved Papers 18th Edition PYQ ... with 100% Detailed Solutions for JEE 2023 Paperback

$~(xiii)~~f(x)=\sin x-\sin 2x~$ in $~~0 \leq x \leq 2\pi.$

Sol. Since $~f(x)~$ is difference of two trigonometric functions , $~f(x)~$ is continuous on $~[0,2\pi].$

Again, $~~f'(x)=\frac{d}{dx}(\sin x-\sin 2x) \\ \text{or,}~~f'(x)=\cos x-2\cos2x \rightarrow(2)$

Clearly, $~f'(x)~$ exists everywhere in $~(0,2\pi).$

Further, $~f(0)=\sin 0-\sin(2 \times 0)=0, \\~~f(2\pi)=\sin \pi-\sin(2 \times \pi)=0 \\ \therefore~~f(0)=f(2\pi).$

So, the given function satisfies all conditions of Rolle's theorem.

$~\therefore~$  there exists at least one value of $~x~$ say $~x=c \in (0,2\pi)~$ such that $~~f'(c)=0 \rightarrow(2)$

So, from $~(1),~(2)~$ we get,

$~~f'(c)=0 \\ \text{or,}~~\cos c-2\cos 2c=0 \\ \text{or,}~~ \cos c-2(2\cos^2c-1)=0 \\ \text{or,}~~ \cos c-4\cos^2c+2=0 \\ \text{or,}~~ 0=4\cos^2c-\cos c-2 \\ \text{or,}~~ \cos c=\frac{-(-1) \pm \sqrt{(-1)^2 - 4 \times 4 \times (-12)}}{2 \times 4} \\ \therefore cos c=\frac{1 \pm\sqrt{33}}{8}$

Now, if $~~x \in [0,2\pi],~~\cos x \in [-1,1].$ So, both values of $~\cos c~$ i.e. $~~\frac{1+\sqrt{33}}{8},~\frac{1-\sqrt{33}}{8}~$ are valid. 

Hence, $~~f'(c)=0 ~~\text{for}~~\\~~c=\cos^{-1}\left(\frac{1+\sqrt{33}}{8}\right)~,~\cos^{-1}\left(\frac{1-\sqrt{33}}{8}\right).$

Hence, there exists  values of $~c~$ in the given interval for which $~~f'(c)=0~~$ and so, Rolle's theorem is verified.