Differentiation (Part-38) | S N De

Differentiation (Part-17) | S N De

 

Differentiation is a fundamental concept in calculus, a branch of mathematics. It refers to the process of finding the rate at which a function changes with respect to its independent variable. In simpler terms, differentiation allows us to determine how a function behaves as its input values (usually denoted as x) change.

The result of differentiation is called the derivative of the function. It provides valuable information about the function's slope or rate of change at any given point. The derivative of a function is itself a function and is often denoted by various notations, such as f'(x), dy/dx, or df/dx, depending on the convention used.

Differentiation is particularly useful for analyzing and solving problems involving rates of change, optimization, and understanding the behavior of functions. It enables us to determine where a function reaches its maximum or minimum values, locate critical points, identify concavity (whether a function is bending upwards or downwards), and analyze the shape of graphs.

The process of differentiation involves applying specific rules and formulas to different types of functions, such as polynomial functions, exponential functions, trigonometric functions, and more. These rules, known as differentiation rules or formulas, allow us to systematically find the derivatives of various functions.

Overall, differentiation plays a crucial role in calculus, providing powerful tools for understanding and analyzing functions and their behavior. It serves as a foundation for many other concepts in mathematics, such as integration, optimization, and differential equations.

9(i)  If $~~x=a(\theta+\sin\theta)~$ and $~~y=a(1+\cos\theta),~$ then find the simplified value of $~~\frac{dy}{dx}~$ and show that, $~~\frac{dy}{dx}=-1,~~$ when $~~\theta=\frac{\pi}{2}.$

Sol.  $~\because~x=a(\theta+\sin\theta) \\ \therefore \frac{dx}{d\theta}=a~\frac{d}{d\theta}(\theta+\sin\theta) \\ \text{or,}~~\frac{dx}{d\theta}\\=a(1+\cos\theta)\\=a \cdot 2\cos^2(\theta/2)\\=2a\cos^2(\theta/2)\rightarrow(1)$

Again, $~~y=a(1+\cos\theta) \\ \therefore \frac{dy}{d\theta}=a~\frac{d}{d\theta}(1+\cos\theta) \\ \text{or,}~~ \frac{dy}{d\theta}\\=a(0-\sin\theta)\\=a \cdot (-2\sin(\theta/2)\cos(\theta/2))\\=-2a\sin(\theta/2)\cos(\theta/2)\rightarrow(2)$

$\therefore~~\frac{dy}{dx}\\=\frac{dy}{d\theta} ~\cdot \frac{d\theta}{dx}\\=\frac{-2a\sin(\theta/2)\cos(\theta/2)}{2a\cos^2(\theta/2)}~~[\text{By (1),(2)}]\\=-\frac{\sin(\theta/2)}{\cos(\theta/2)}\\=-\tan(\theta/2)\\ \therefore ~~\left[\frac{dy}{dx}\right]_{\theta=\pi/2}\\=[-\tan(\theta/2)]_{\theta=\pi/2}\\=-\tan\left(\frac{\pi}{2 \cdot 2}\right)\\=-\tan(\pi/4)\\=-1~~\text{(showed)}$

(ii)$~x=\frac{\sin^3t}{\sqrt{\cos2t}}~$ and $~y=\frac{\cos^3t}{\sqrt{\cos2t}},~~$ show that $~~\frac{dy}{dx}=0~$ at $~t=\pi/6.$

Sol. $~~\because x=\frac{\sin^3t}{\sqrt{\cos2t}} \\ \therefore \frac{dx}{dt}\\=\frac{d}{dt}\left(\frac{\sin^3t}{\sqrt{\cos2t}}\right)\\=\frac{\sqrt{\cos2t}\cdot\frac{d}{dt}(\sin^3t)-\sin^3t\frac{d}{dt}(\sqrt{\cos2t})}{(\sqrt{\cos2t})^2}\\=\frac{\sqrt{\cos2t}\cdot 3\sin^2t\cos t-\sin^3t\cdot \frac{1}{2\sqrt{\cos2t}}(-2\sin2t)}{\cos2t}\\=\frac{3\cos2t \cdot \sin^2t\cdot \cos t+\sin^3t \cdot \sin2t}{\sqrt{\cos2t}\cdot\cos2t}\\=\frac{3\cos2t \cdot \sin^2t\cdot \cos t+2\sin^4t \cdot \cos t}{\sqrt{\cos2t}\cdot\cos2t}\\=\frac{\sin^2t\cdot \cos t(3\cos 2t+2\sin^2t)}{\sqrt{\cos2t}\cdot \cos 2t}\rightarrow(1)$

Again, $~y=\frac{\cos^3t}{\sqrt{\cos2t}} \\ \therefore \frac{dy}{dt}\\=\frac{d}{dt}\left(\frac{\cos^3t}{\sqrt{\cos2t}}\right)\\=\frac{\sqrt{\cos2t}~\frac{d}{dt}(\cos^3t)-\cos^3t\cdot \frac{d}{dt}(\sqrt{\cos2t})}{(\sqrt{\cos2t})^2}\\=\frac{\sqrt{\cos2t}\cdot 3\cos^2t(-\sin t)-\cos^3t \cdot \frac{1}{2\sqrt{\cos2t}}\cdot (-2\sin2t)}{\cos2t}\\=\frac{-3\cos2t \cdot \cos^2t \cdot \sin t+\cos^3t \cdot \sin2t }{\sqrt{\cos2t}\cos2t}\\=\frac{-3\cos2t \cdot \cos^2t \cdot \sin t+ 2\sin t \cdot \cos^4t}{b}\\=\frac{\sin t\cdot \cos^2t(2\cos^2t-3\cos2t)}{b}\rightarrow(2) \\ \therefore \frac{dy}{dx}\\=\frac{dy}{dt} \cdot \frac{dt}{dx}\\=\frac{\sin t \cdot \cos^2t(2\cos^2t-3\cos2t)}{\sin^2t \cdot \cos t(3\cos2t+2\sin^2t)}~~[\text{By (1),(2)}]\\=\frac{\cos t(2\cos^2t-3\cos2t)}{\sin t(3\cos 2t+2\sin^2t)} \\ \therefore ~\left[\frac{dy}{dx}\right]_{t=\pi/6}\\=\frac{\cos(\pi/6)(2\cos^2(\pi/6)-3\cos(\pi/3))}{\sin(\pi/6)(3\cos(\pi/3)+2\sin^2(\pi/6))}\\=\frac{\frac{\sqrt 3}{2}\left(2 \times \frac 34-\frac 32\right)}{\frac 12\left(3 \times \frac 12+2 \times \frac 14\right)}\\=0~~\text{(showed)}$

Find the differential coefficients of $\,[~10-13~]:$

10.$~\frac{\tan^{-1}x}{1+\tan^{-1}x}~$ w.r.t. $~~\tan^{-1}x.$

Sol. let $~~v=\frac{\tan^{-1}x}{1+\tan^{-1}x}\rightarrow(1),\\~~u=\tan^{-1}x \rightarrow(2).$

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

$~~~~v=\frac{u}{1+u} \\ \therefore \frac{dv}{du}\\=\frac{d}{du}\left(\frac{u}{1+u}\right)\\=\frac{(1+u)\frac{d}{du}(u)-u\frac{d}{du}(1+u)}{(1+u)^2}\\=\frac{(1+u) \cdot 1-u \cdot 1}{(1+u)^2}\\=\frac{1}{(1+u)^2}\\=\frac{1}{(1+\tan^{-1}x)^2}~~~\text{(ans.)}$ 

11.$~\tan^{-1}\frac{\sqrt{1+x^2}-1}{x}~$ w.r.t. $~~\tan^{-1}x.$

Sol. let $~~v=\tan^{-1}\frac{\sqrt{1+x^2}-1}{x} \rightarrow(1),\\~~u=\tan^{-1}x \\ \Rightarrow x=\tan u\rightarrow(2)$

So, from $\,(1)\,$ and $\,(2)\,$ we get,

$~~~~~v\\=\tan^{-1}\left(\frac{\sqrt{1+\tan^2u}-1}{\tan u}\right)\\=\tan^{-1}\left(\frac{\sqrt{\sec^2u}-1}{\tan u}\right)\\=\tan^{-1}\left(\frac{\sec u-1}{\tan u}\right)\\=\tan^{-1}\left(\frac{\frac{1}{\cos u}-1}{\frac{\sin u}{\cos u}}\right)\\=\tan^{-1}\left(\frac{1-\cos u}{\sin u}\right)\\=\tan^{-1}\left(\frac{2\sin^2(u/2)}{2\sin(u/2)\cos(u/2)}\right)\\=\tan^{-1}\left(\frac{\sin(u/2)}{\cos(u/2)}\right)\\=\tan^{-1}\left(\tan(u/2)\right)\\=\frac u2 \\ \therefore v=\frac u2 \\ \text{or,}~~\frac{dv}{du}=\frac 12~\frac{d}{du}(u) \\ \therefore \frac{dv}{du}=\frac 12~~\text{(ans.)}$

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$~12.~~\cos^{-1}\frac{1-x^2}{1+x^2}~$ w.r.t. $~~\sin^{-1}\frac{2x}{1+x^2}.$

Sol. let $~~v=\cos^{-1}\frac{1-x^2}{1+x^2},\\~~~u=\sin^{-1}\frac{2x}{1+x^2}$

We put, $~~x=\tan\theta~~$ so that 

$~~v\\=\cos^{-1}\left(\frac{1-\tan^2\theta}{1+\tan^2\theta}\right)\\=\cos^{-1}\left(\frac{\cos^2\theta-\sin^2\theta}{\cos^2\theta+\sin^2\theta}\right)\\=\cos^{-1}\left(\frac{\cos2\theta}{1}\right)\\=\cos^{-1}(\cos2\theta)\\=2\theta\rightarrow(1)$

Similarly, $~u\\=\sin^{-1}\left(\frac{2\tan\theta}{1+\tan^2\theta}\right)\\=\sin^{-1}\left(\frac{2\tan\theta}{\sec^2\theta}\right)\\=\sin^{-1}\left(2\cdot \frac{\sin \theta}{\cos\theta} \cdot \cos^2\theta\right)\\=\sin^{-1}(2\sin\theta\cos\theta)\\=\sin^{-1}(\sin2\theta)\\=2\theta\rightarrow(2)$

Hence, from $\,(1),(2)~$ we get,

$~~v=u \\ \therefore \frac{dv}{du}=\frac{d}{du}(u)=1~~\text{(ans.)}$

13.$~x^{\sin^{-1}x}~~$ w.r.t. $~~\sin^{-1}x.$

Sol. $~~v=x^{\sin^{-1}x}\rightarrow(1),\\~~~~~u=\sin^{-1}x\\ \Rightarrow x=\sin u \rightarrow(2)$

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

$~~~~~~\log v=\log(x^{\sin^{-1}x}) \\ \text{or,}~~\log v=\sin^{-1}x \log x \\ \text{or,}~~\log v=u\log(\sin u)~~[\text{By (2)}] \\ \therefore~\frac 1v~\frac{dv}{du}=\frac{d}{du}(u) \cdot \log(\sin u)\\~~~~+u \cdot \frac{d}{du}(\log(\sin u)) \\ \text{or,}~~ \frac 1v~\frac{dv}{du}=1 \cdot \log(\sin u)+u \cdot \frac{1}{\sin u}\cdot \cos u \\ \text{or,}~~ \frac{dv}{du}=v\left[\log(\sin u)+\frac{u}{\sin u} \cdot \sqrt{1-\sin^2u}\right] \\ \text{or,}~~ \frac{dv}{du}=x^{\sin^{-1}x}\left[\log x+\frac{\sin^{-1}x}{x} \cdot \sqrt{1-x^2}\right]~~\text{(ans.)}$