Differentiation (Part-38) | S N De

Differentiation (Part-1) | S N De

 

In mathematics, differentiation is a fundamental operation in calculus that involves finding the rate at which a function changes with respect to its independent variable. It allows us to determine how a function behaves as its input values (often denoted as x) vary.

The result of differentiation is called the derivative of the function. The derivative represents the instantaneous rate of change of the function at any given point. It provides valuable information about the slope, or steepness, of the function's graph at different points.

The process of differentiation involves applying specific rules and formulas to different types of functions. These rules, known as differentiation rules or formulas, enable us to systematically find the derivatives of various functions. Examples of differentiation rules include the power rule, product rule, quotient rule, and chain rule, among others.

The derivative of a function can be denoted in different notations, such as f'(x), dy/dx, or df/dx, depending on the convention used. The derivative is itself a function and can be evaluated at specific points to obtain the slope of the tangent line to the graph of the function at those points.

Differentiation has numerous applications in mathematics and other fields. It helps us analyze the behavior of functions, determine increasing or decreasing intervals, find maximum and minimum values, solve optimization problems, and study rates of change in various contexts. Differentiation is also a key component in integral calculus, where it is used to calculate areas under curves and solve differential equations.

Overall, differentiation is a powerful tool in calculus that allows us to understand and quantify the changes and rates of change in functions. It provides a deeper understanding of mathematical functions and their behavior, making it a fundamental concept in mathematical analysis and applications.

1. Find $~\frac{dy}{dx}~$ using the rule $~\frac{dy}{dx}=\frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dx}$

$~(i)~~y=\sin\sqrt{x^2+a^2}$

Sol. let $~y=\sin u \Rightarrow \frac{dy}{du}=\cos u $

Again, let $~~u=\sqrt v \Rightarrow \frac{du}{dv}=\frac{1}{2\sqrt v}$

Here, $~~v=x^2+a^2 \Rightarrow \frac{dv}{dx}=2x$

So, $~~\frac{dy}{dx}=\frac{dy}{du} .\frac{du}{dv}. \frac{dv}{dx}\\ \text{or,}~~ \frac{dy}{dx}=\cos u .\frac{1}{2\sqrt v}.~2x\\~~~~~~~~~~~=\cos\sqrt{x^2+a^2} \cdot \frac{x}{\sqrt{x^2+a^2}}\\~~~~~~~~~~~=\frac{x \cos{\sqrt{x^2+a^2}}}{\sqrt{x^2+a^2}}~~\text{(ans.)}$

$~(ii)~~y=\log_2(\sin x^3)$

Sol. $~~~y=\log_2(\sin x^3)=\log_e(\sin x^3) \times \log_e2$

let $~~~~y=\log_eu \times \log_2e\\~~~[\text{where}~~u=\sin x^3] \\ \therefore \frac{dy}{du}=\log_2e \cdot \frac 1u $

Now, $~~~~u=\sin v~~~[\text{where}~~v=x^3]\\ \Rightarrow \frac{du}{dv}=\cos v$

$~~v=x^3 \Rightarrow \frac{dv}{dx}=3x^2 \\ \therefore \frac{dy}{dx}=\frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dx} \\ \text{or,}~~ \frac{dy}{dx}=\log_2e \cdot \frac 1u \cdot \cos v \cdot 3x^2 \\ \text{or,}~~ \frac{dy}{dx}=\log_2e \cdot \frac{\cos v}{\sin v}\cdot3x^2 \\ \therefore \frac{dy}{dx}=3x^2\cot(x^3)\log_2e~~\text{(ans.)}$

$~(iii)~~y=e^{\sqrt{\cos x}}$

Sol. $~~y=e^{\sqrt{\cos x}} \\ \text{or,}~~y=e^u ~~[~\text{where}~~u=\sqrt{\cos x}]\\ \text{or,}~~\frac{dy}{du}=e^u $

Again, $~u=\sqrt v \Rightarrow \frac{du}{dv}=\frac{1}{2\sqrt v}$

Now, $~v=\cos x \Rightarrow \frac{dv}{dx}=-\sin x$

So, $~~\frac{dy}{dx}=\frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dx}\\ \text{or,}~~\frac{dy}{dx}=e^u \cdot \frac{1}{2\sqrt v} \cdot (-\sin x) \\ \text{or,}~~\frac{dy}{dx}=-\frac{e^{\sqrt{\cos x}}\cdot \sin x}{2\sqrt{\cos x}}~~\text{(ans.)}$

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$~(iv)~~y=\frac{1}{\sqrt{\log \sec x}}$

Sol. $~~y=\frac{1}{\sqrt{\log \sec x}} \\ \text{or,}~~y=\frac 1u~~[~u=\sqrt{\log \sec x}] \\ \text{or,}~~\frac{dy}{du}=-\frac{1}{u^2}$

Again, $~~u=\sqrt v \\ \text{or,}~~\frac{du}{dv}=\frac{1}{2\sqrt v}$

$~~v=\log(\sec x) \\ \text{or,}~~\frac{dv}{dx}=\frac{1}{\sec x} \cdot \sec x \tan x \\ \text{or,}~~\frac{dv}{dx}=\tan x \\ \therefore \frac{dy}{dx}=\frac{dy}{du} .\frac{du}{dv}.\frac{dv}{dx} \\ \text{or,}~~\frac{dy}{dx}=(-1/u^2)\cdot \frac{1}{2\sqrt v} \cdot \tan x \\ \text{or,}~~ \frac{dy}{dx}=-\frac{\tan x}{2\log(\sec x) \cdot\sqrt{\log(\sec x)}}\\ \text{or,}~~\frac{dy}{dx}=-\frac 12 \tan x[\log(\sec x)]^{-3/2}~~\text{(ans.)}$

$~(v)~~y=\cos^{-1}\sqrt{2x-3}$

Sol. let $~~y=\cos^{-1}u\Rightarrow \frac{dy}{du}=-\frac{1}{\sqrt{1-u^2}}\\~~~[~\text{where}~~u=\sqrt{2x-3}~~]$

Again, suppose $u=\sqrt v \Rightarrow \frac{du}{dv}=\frac{1}{2\sqrt v}$

$~~v=2x-3 \Rightarrow \frac{dv}{dx}=2 \\ \therefore \frac{dy}{dx}=\frac{dy}{du} .\frac{du}{dv}. \frac{dv}{dx}\\ \text{or,}~~ \frac{dy}{dx}=-\frac{1}{\sqrt{1-u^2}} \cdot \frac{1}{2\sqrt v} \cdot 2 \\ \text{or,}~~\frac{dy}{dx}=-\frac{1}{\sqrt{1-(2x-3)}} \cdot \frac{1}{\sqrt{2x-3}} \\ \text{or,}~~\frac{dy}{dx}=-\frac{1}{\sqrt{4-2x}\sqrt{2x-3}} \\ \text{or,}~~\frac{dy}{dx}=-\frac{1}{\sqrt 2} \cdot \frac{1}{\sqrt{2-x}\sqrt{2x-3}}~~\text{(ans.)}$

2.  If the inverse function of $~~y=f(x)~$ is $~~x=g(y)~$ and $~f'(x)=\frac{1}{1+x^2},~$ then prove that, $~~g'(x)=1+[g(x)]^2.$

Sol. $~f'(x)=\frac{1}{1+x^2}\\ \text{or,}~~ f(x)=\displaystyle\int{\frac{dx}{1+x^2}}=\tan^{-1}x \\ \therefore y=\tan^{-1}x~~[\because y=f(x)] \\ \Rightarrow x=\tan y \\ \Rightarrow g(y)=\tan y\rightarrow(1)\\~~[\because x=g(y)]$ 

Hence, from $\,(1)\,$ we can say, 

$~g(x)=\tan x ~~[~\text{replacing}~ ~y~~\text{by}~~ x] \\ \therefore g'(x)=\frac{d}{dx}(\tan x)=\sec^2x\rightarrow(2)$

Again, $~~~~1+[g(x)]^2\\=1+\tan^2x\\=\sec^2x \rightarrow(3)$

So, from $\,(2)\,$ and $\,(3)\,$ we can conclude that 

$~~g'(x)=1+[g(x)]^2~~~\text{(proved)}$