Differentiation (Part-38) | S N De

 $~12.~$ If $~f(x)=\left(\frac{a+x}{1+x}\right)^{a+1+2x},~$ show that, $~~f'(0)=a^{a+1}\left[2\log a+\frac{1-a^2}{a}\right]$

Sol. $~f(x)=\left(\frac{a+x}{1+x}\right)^{a+1+2x}\rightarrow(1) \\ \text{or,}~~  \log f(x)=\log  \left(\frac{a+x}{1+x}\right)^{a+1+2x}\\ \text{or,}~~ \log f(x)= (a+1+2x)\log \left(\frac{a+x}{1+x}\right)\\ \therefore \frac{1}{f(x)} \cdot f'(x)=2 \log\left(\frac{a+x}{1+x}\right)\\~~~~+(a+1+2x) \cdot \frac{1}{\left(\frac{a+x}{1+x}\right)}\cdot \frac{d}{dx} \left(\frac{a+x}{1+x}\right) \\ \text{or,}~~  \frac{f'(x)}{f(x)}=2\log \left(\frac{a+x}{1+x}\right) \\~~~~+\frac{(a+1+2x)(1+x)}{a+x} \cdot  \frac{(1+x)\cdot 1-(a+x) \cdot 1}{(1+x)^2}\\ \text{or,}~~ \frac{f'(x)}{f(x)}=2\log \left(\frac{a+x}{1+x}\right)\\~~~~+\frac{(a+1+2x)(1+x)}{a+x} \cdot \frac{1+x-a-x}{(1+x)^2}\rightarrow(2)$ 

At $~x=0,~~f(0)=a^{a+1}~[~\text{By (1)}]$

So, at $~x=0,~$ we get from $\,(2),$

$~~\frac{f'(0)}{f(0)}=2 \log \left(\frac{a+0}{1+0}\right)+\frac{(a+1)}{a} \cdot \frac{1-a}{(1+0)^2} \\ \therefore f'(0)=f(0)\left[2\log a+\frac{(1+a)(1-a)}{a}\right] \\ \text{or,}~~f'(0)=a^{a+1}\left[2 \log a+\frac{1-a^2}{a}\right]~~~\text{(proved)}$

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$\,13.~$ If $~f(x)=\left(\frac{a+x}{b+x}\right)^x+\cos x,~~$ find $~f'(0).$

Sol. $~f(x)=\left(\frac{a+x}{b+x}\right)^x+\cos x \\ \text{or,}~~  f(x)=u+v~~\text{(say)} \\ \therefore f'(x)=\frac{du}{dx}+\frac{dv}{dx} \rightarrow(1) $

Now, $~~u=\left(\frac{a+x}{b+x}\right)^x \\ \text{or,}~~ \log u= \log \left(\frac{a+x}{b+x}\right)^x \\ \text{or,}~~  \log u=x \log\left(\frac{a+x}{b+x}\right) \\ \therefore \frac 1u \frac{du}{dx}=\log \left(\frac{a+x}{b+x}\right)\\~~~~~~ +\frac{x}{\left(\frac{a+x}{b+x}\right)} \cdot \frac{d}{dx} \left(\frac{a+x}{b+x}\right)\\ \text{or,}~~ \frac 1u \frac{du}{dx}=\log \left(\frac{a+x}{b+x}\right)\\~~~~~~+\frac{x(b+x)}{a+x} \times \frac{(b+x) \cdot 1-(a+x)\cdot 1}{(b+x)^2} \\ \text{or,}~~  \frac 1u \frac{du}{dx}= \log \left(\frac{a+x}{b+x}\right)\\~~~~~~+\frac{x}{(a+x)(b+x)} \cdot (b-a) \\ \text{or,}~~  \frac{du}{dx}=\left(\frac{a+x}{b+x}\right)^x \left[\log \left(\frac{a+x}{b+x}\right)\\~~~~~~+\frac{x}{(a+x)(b+x)} \cdot (b-a)\right] \\ \text{or,}~~  \left[\frac{du}{dx}\right]_{x=0}=\left(\frac ab\right)^0 \left[\log\left(\frac ab\right)+0\right] \\ \text{or,}~~ \left[\frac{du}{dx}\right]_{x=0}=\log\left(\frac ab\right) \rightarrow(2)$ 

Similarly, $~~v=\cos x \\ \text{or,}~~ \frac{dv}{dx}=-\sin x \\ \text{or,}~~ \left[\frac{dv}{dx}\right]_{x=0}=0 \rightarrow(3)$

Hence, from $\,(1),\, (2),\,(3)\,$ we get,

$~~f'(0)=\left[\frac{du}{dx}\right]_{x=0}+\left[\frac{dv}{dx}\right]_{x=0} \\ \text{or,}~~ f'(0)=\log\left(\frac ab\right)+0 \\ \therefore f'(0)=\log\left(\frac ab\right)~~\text{(ans.)}$

$\,14.\,$ If $~y=\tan^{-1}\left(\frac{\sqrt{1+t^2}+\sqrt{1-t^2}}{\sqrt{1+t^2}-\sqrt{1-t^2}}\right),~$ find the value of $\,\frac{dy}{dt}.$

Sol. $~y=\tan^{-1}\left(\frac{\sqrt{1+t^2}+\sqrt{1-t^2}}{\sqrt{1+t^2}-\sqrt{1-t^2}}\right)\rightarrow(1)$

let $~t^2=\cos 2\theta~$ so that 

$~~y=\tan^{-1} \left(\frac{\sqrt{1+\cos 2\theta}+\sqrt{1-\cos2\theta}}{\sqrt{1+\cos2\theta}-\sqrt{1-\cos 2\theta}}\right) ~~[~\text{By (1)}]  \\ \text{or,}~~y=\tan^{-1} \left(\frac{\sqrt{2\cos^2\theta}+\sqrt{2\sin^2\theta}}{\sqrt{2\cos^2\theta}-\sqrt{2\sin^2\theta}}\right) \\ \text{or,}~~ y=\tan^{-1} \left(\frac{\sqrt 2(\cos \theta+\sin \theta))}{\sqrt 2(\cos \theta-\sin \theta)}\right) \\ \text{or,}~~ y=\tan^{-1} \left(\frac{\cos \theta+\sin \theta}{\cos \theta-\sin \theta}\right)\\ \text{or,}~~ y=\tan^{-1} \left(\frac{1+\tan \theta}{1-\tan \theta}\right) \\ \text{or,}~~y=\tan^{-1} \left(\frac{\tan (\pi/4)+\tan \theta}{1-\tan(\pi/4) \tan \theta}\right) \\ \text{or,}~~y=\tan^{-1}[\tan (\pi/4+\theta)] \\ \text{or,}~~y=\frac{\pi}{4}+\theta\\~~ \therefore ~~y=\frac{\pi}{4}+\frac 12 \cos^{-1}t^2  \\ \text{or,}~~\frac{dy}{dt}=\frac 12 \left[-\frac{1}{\sqrt{1-(t^2)^2}} \times (2t)\right] \\~\therefore~~ \frac{dy}{dt}=-\frac{t}{\sqrt{1-t^4}}~~\text{(ans.)}$

$~15.~~y=\cot^{-1} \left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right),~~$ show that , $\left[\frac{dy}{dx}\right]_{x=1/2}=-\frac{1}{\sqrt 3}.$

Sol. $~y=\cot^{-1} \left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)\rightarrow(1)$

Let $~x=\cos2\theta~$ so that 

$~~y=\cot^{-1}\left(\frac{\sqrt{1+\cos2\theta}-\sqrt{1-\cos2\theta}}{\sqrt{1+\cos2\theta}+\sqrt{1-\cos2\theta}}\right)~~[\text{By (1)}] \\ \text{or,}~~y=\tan^{-1} \left(\frac{\sqrt{1+\cos2\theta}+\sqrt{1-\cos2\theta}}{\sqrt{1+\cos2\theta}-\sqrt{1-\cos2\theta}}\right) \\ \text{or,}~~y=\tan^{-1} \left(\frac{\sqrt{2\cos^2\theta}+\sqrt{2\sin^2\theta}}{\sqrt{2\cos^2\theta}-\sqrt{2\sin^2\theta}}\right) \\ \text{or,}~~ y=\tan^{-1} \left[\frac{\sqrt 2(\cos \theta+\sin\theta)}{\sqrt 2(\cos \theta-\sin \theta)}\right] \\ \text{or,}~~ y =\tan^{-1}\left[\frac{\cos \theta+\sin\theta}{\cos \theta-\sin\theta}\right] \\ \text{or,}~~ y=\tan^{-1} \left(\frac{1+\tan\theta}{1-\tan\theta}\right) \\ \text{or,}~~ y=\tan^{-1} \left(\frac{\tan (\pi/4)+\tan \theta}{1-\tan(\pi/4) \tan \theta}\right) \\ \text{or,}~~ y=\tan^{-1} \left[\tan\left(\frac{\pi}{4}+\theta\right)\right] \\ \text{or,}~~ y=\frac{\pi}{4}+ \theta \\ \text{or,}~~ y=\frac{\pi}{4}+\frac 12 \cos^{-1}x  \\ \therefore ~\frac{dy}{dx}=-\frac 12 \cdot \frac{1}{\sqrt{1-x^2}} \\ \text{so,}~~\left[\frac{dy}{dx}\right]_{x=1/2}=-\frac 12 \cdot \frac{1}{\sqrt{1-\frac 14}}=-\frac{1}{\sqrt 3}~~\text{(proved)}$

$\,16.\,$ If $~(1+x)^n=C_0+C_1x+C_2x^2+\cdots+C_nx^n,~$ prove that,

$~~(i)~C_0+2C_1+3C_2\\~~~~~~~~+\cdots+(n+1)C_n=(n+2) \cdot 2^{n-1}\\ ~(ii)~C_0-2C_1+3C_2\\~~~~~~~~+\cdots+(-1)^{n-1}\cdot nC_n=0$

Sol. $~(1+x)^n=C_0+C_1x+C_2x^2+\cdots+C_nx^n\rightarrow(1) \\ \therefore x(1+x)^n=C_0x+C_1x^2+C_2x^3\\~~~~~~~~+\cdots +C_nx^{n+1}\rightarrow(2)$

Now, differentiating $\,(2)\,$ w.r.t $\,x,$ we get,

$~~(1+x)^n+x \cdot n(1+x)^{n-1}\\=C_0+2C_1x+3C_2x^2+\cdots+(n+1)C_nx^n \\ \therefore (1+1)^n+1\cdot n(1+1)^{n-1}\\~~~~~~~~=C_0+2C_1+3C_2+\cdots +(n+1)C_n\\~~~[\text{Putting }~~x=1] \\ \text{or,}~~ 2^n+n\cdot 2^{n-1}=C_0+2C_1+3C_2\\~~~~~~~~+\cdots +(n+1)C_n \\ \therefore C_0+2C_1+3C_2\\~~~~~~~~+\cdots +(n+1)C_n=2^{n-1}(2+n) \\ \text{so,}~~C_0+2C_1+3C_2\\~~~~~~~~+\cdots +(n+1)C_n=(n+2)\cdot 2^{n-1}\\~~~~~~~~~~\text{(proved)}$

$~2~$nd part:

Sol.  $~(1+x)^n=C_0+C_1x+C_2x^2\\~~~~~~~~~~~+\cdots+C_nx^n\rightarrow(1) $

Differentiating $\,(1)\,$ w.r.t. $\,x,\,$ we get,

$~~~n(1+x)^{n-1}=C_1+2C_2x+\cdots+nx^{n-1}\cdot C_n \\ \text{Now, putting}~x=-1,~~\text{we get}, \\ ~~n(1-1)^{n-1}=C_0-2C_1+3C_2\\~~~~~~~~~~~+\cdots+(-1)^{n-1}\cdot nC_n \\ \therefore~~C_0-2C_1+3C_2\\~~~~~~~~~~~+\cdots+(-1)^{n-1}\cdot nC_n=0~~\text{(proved)}$

$~17(i)~~$ If $~\sin y=x\sin(a+y),~$ prove that, $~\frac{dy}{dx}=\frac{\sin a}{1-2x\cos a+x^2}$

Sol. $~~~~~~\sin y=x\sin(a+y) \\ \text{or,}~~\sin y=x (\sin a \cos y+\cos a\sin y) \\ \text{or,}~~\sin y-x \cos a \sin y=x \sin a\cos y \\ \text{or,}~~ \sin y(1-x \cos a)=x \sin a \cos y \\ \text{or,}~~ \frac{\sin y}{\cos y}=\frac{x \sin a}{1-x\cos a} \\ \text{or,}~~ \tan y=\frac{x \sin a}{1-x\cos a} \\ \text{or,}~~ y=\tan^{-1} \left(\frac{x \sin a}{1-x\cos a} \right) \\ \therefore ~\frac{dy}{dx}\\=\frac{1}{1+\left(\frac{x \sin a}{1-x\cos a}\right)^2} \cdot \frac{d}{dx}\left(\frac{x \sin a}{1-x\cos a}\right)\\=\frac{(1-x\cos a)^2}{(1-x\cos a)^2+(x\sin a)^2}\\~~~~~~ \times \frac{\sin a(1-x\cos a)-x\sin a(-\cos a)}{(1-x \cos a)^2}\\=\frac{\sin a(1-x\cos a)-x\sin a(-\cos a)}{(1-x\cos a)^2+(x\sin a)^2}\\=\frac{\sin a-x \sin a\cos a+x \sin a \cos a}{1-2x\cos a+x^2\cos^2a+x^2\sin^2a}\\=\frac{\sin a}{1-2x\cos a+x^2(\cos^2a+\sin^2a)}\\=\frac{\sin a}{1-2x\cos a+x^2}~~~\text{(proved)}$ 

#$~17(ii)~~$ If $~y=\begin{vmatrix} f(x) & g(x) & h(x)\\ l& m &n \\ a &b &c \end{vmatrix},~$ prove that , $~~\frac{dy}{dx}=\begin{vmatrix} f'(x) & g'(x) & h'(x)\\ l& m &n \\ a &b &c \end{vmatrix}$ 

 Sol. $~~~~~~y=\begin{vmatrix} f(x) & g(x) & h(x)\\ l& m &n \\ a &b &c \end{vmatrix} \\ \text{or,}~~ ~y=f(x) (mc-nb)-g(x)(lc-na)\\~~~+h(x)(lb-am) \\ \therefore\frac{dy}{dx}=f'(x)(mc-nb)-g'(x)(lc-na)\\~~~+h'(x)(lb-am) \\ \text{or,}~~\frac{dy}{dx}=f'(x)\begin{vmatrix} m & n\\ b & c \end{vmatrix}-g'(x) \begin{vmatrix} l & n\\ a & c \end{vmatrix}\\~~~+h'(x)\begin{vmatrix} l & m\\ a & b \end{vmatrix} \\ \therefore ~~\frac{dy}{dx}=\begin{vmatrix} f'(x) &g'(x) &h'(x) \\ l &m &n \\ a &b &c \end{vmatrix}~~\text{(proved)}$