Differentiation (Part-38) | S N De

 Differentiation (Part-8) | S N De

 

$~18.~$ If $~x=\frac{t-e^{-t^2}}{2t^2},~$ prove that, $~t~\frac{dx}{dt}+2x=\frac{1}{2t}+e^{-t^2}.$

Sol.  $~~~~~~~x=\frac{t-e^{-t^2}}{2t^2} \\ \text{or,}~~  x=\frac{1}{2t}-\frac{e^{-t^2}}{2t^2} \\ \therefore \frac{dx}{dt}=-\frac{1}{2t^2}-\frac{-e^{-t^2}.2t.2t^2-4t.e^{-t^2}}{(2t^2)^2} \\ \text{or,}~~  \frac{dx}{dt}=\frac{4t^3e^{-t^2}+4te^{-t^2}}{4t^4}-\frac{1}{2t^2} \\ \text{or,}~~  \frac{dx}{dt}=\frac{4te^{-t^2}(t^2+1)}{4t^4}-\frac{1}{2t^2} \\ \text{or,}~~  \frac{dx}{dt}=\frac{e^{-t^2}(t^2+1)}{t^3}-\frac{1}{2t^2} \\ \text{or,}~~  t~\frac{dx}{dt}=\frac{e^{-t^2}(t^2+1)}{t^2}-\frac{1}{2t} \\ \text{or,}~~  t~\frac{dx}{dt}+2x\\=\frac{e^{-t^2}(t^2+1)}{t^2}-\frac{1}{2t}+2\left(\frac{1}{2t}-\frac{e^{-t^2}}{2t^2}\right)\\=\frac{e^{-t^2}(t^2+1)}{t^2}-\frac{e^{-t^2}}{t^2}+ \left(\frac{2}{2t}-\frac{1}{2t}\right)\\=\frac{e^{-t^2}t^2+e^{-t^2}-e^{-t^2}}{t^2}+\frac{1}{2t}\\=e^{-t^2}+\frac{1}{2t}~~\text{(proved)}$

$~19.~$ If $~y\sqrt{x^2+1}=\log(\sqrt{x^2+1}-x)~$, show that, $~(x^2+1)~\frac{dy}{dx}+xy+1=0.$

Sol. $~~~~~y\sqrt{x^2+1}=\log(\sqrt{x^2+1}-x) \rightarrow(1)\\ \text{or,}~~y^2(\sqrt{x^2+1})^2=[\log(\sqrt{x^2+1}-x)]^2 \\ \text{or,}~~  y^2(x^2+1)=[\log(\sqrt{x^2+1}-x)]^2 \\ \therefore 2y(x^2+1)~\frac{dy}{dx}+y^2 \cdot 2x\\=2\log(\sqrt{x^2+1}-x)\\~~~~\cdot\frac{d}{dx}~[\log(\sqrt{x^2+1}-x)] \\=2\log(\sqrt{x^2+1}-x)\\~~~~\cdot \frac{1}{\sqrt{x^2+1}-x} \cdot \left(\frac 12 \cdot \frac{1}{\sqrt{x^2+1}} \cdot 2x -1\right)\\=2\log(\sqrt{x^2+1}-x)\\~~~~\cdot \frac{1}{\sqrt{x^2+1}-x}\cdot \frac{-(\sqrt{x^2+1}-x)}{\sqrt{x^2+1}} \\=\frac{-2\log(\sqrt{x^2+1}-x)}{\sqrt{x^2+1}}\\=-2y~~~[\text{By (1)}] \\ \therefore 2y\left[(x^2+1)~\frac{dy}{dx}+xy\right]=-2y \\ \text{or,}~~(x^2+1)~\frac{dy}{dx}+xy=-1 \\ \text{or,}~~(x^2+1)~\frac{dy}{dx}+xy+1=0~~\text{(showed)}$

$~20.~~$ Using the method of differentiation deduce the formula of :

$~(i)~~\cos(x+\alpha)~$ starting from the formula of $~~\sin(x+\alpha).$

Sol. $\,\,\,\sin(x+\alpha)=\sin x \cos \alpha+\cos x\sin \alpha \\ \therefore~\frac{d}{dx}\left(\sin(x+\alpha)\right)=\cos \alpha \cdot \frac{d}{dx}(\sin x)\\~~~~~~~~~~~~~~~~~~~+\sin \alpha \cdot \frac{d}{dx}(\cos x) \\ \text{or,}~~\cos(x+\alpha)=\cos \alpha\cos x-\sin\alpha \sin x ~~\text{(ans.)}$

$~20.~~$ Using the method of differentiation deduce the formula of :

$~(ii)~~\sin(x+\alpha)~$ starting from the formula of $~~\cos(x+\alpha).$

Sol. $\,\,\,\cos(x+\alpha)=\cos \alpha\cos x-\sin\alpha \sin x \\ \therefore~\frac{d}{dx}\left(\cos(x+\alpha)\right)=\cos \alpha \cdot \frac{d}{dx}(\cos x)\\~~~~~~~~~~~~~~-\sin \alpha \cdot \frac{d}{dx}(\sin x) \\ \text{or,}~~-\sin(x+\alpha)=-\cos \alpha\sin x \\~~~~~~~~~~~~~~-\sin\alpha \cos x \\ \text{or,}~~ \sin(x+\alpha)=\sin x \cos \alpha+\cos x \sin \alpha\text{(ans.)}$

$~20(iii)~~\sin 2x~~$ starting from the formula of $~~\cot 2x.$

Sol. We know, $~~\cot(A+B)=\frac{\cot A \cot B-1}{\cot B+\cot A}\rightarrow(1)$

Let $~~A=B=x~~\text{so that from }~(1)~$ we get,

$~~\cot(x+x)=\frac{\cot x \cdot \cot x-1}{\cot x+\cot x} \\ \therefore \cot 2x=\frac{\cot^2x-1}{2\cot x} \\ \text{or,}~~  \cot 2x=\frac{1-\tan^2x}{2\tan x} \\ \therefore \cot 2x=\frac 12 \cot x-\frac 12 \tan x \\~~~~~~ \text{Differentiating we get,}~~\\ ~~~-2\csc^22x=-\frac 12\csc^2x-\frac 12\sec^2x \\ \text{or,}~~ 4 \csc^22x=\frac{1}{\sin^2x}+\frac{1}{\cos^2x}\\ \text{or,}~~  4\csc^22x =\frac{\cos^2x+\sin^2x}{\sin^2x\cos^2x} \\ \text{or,}~~  \csc^22x=\frac{1}{(2\sin x\cos x)^2} \\ \text{or,}~~  \sin^22x =(2\sin x\cos x)^2 \\ ~~\therefore ~~\sin 2x=2\sin x\cos x~~\text{(ans.)}$

$~21.~$ If $~y=1+\frac{a_1}{x-a_1}+\frac{a_2x}{(x-a_1)(x-a_2)} \\~~~~~~~~~~+\frac{a_3x^2}{(x-a_1)(x-a_2)(x-a_3)},~~$ 

show that, $~~\frac{dy}{dx}=\frac yx \left[\frac{a_1}{a_1-x}+\frac{a_2}{a_2-x}+\frac{a_3}{a_3-x}\right]$

Sol.$~~~~~y=\left(1+\frac{a_1}{x-a_1}\right)+\frac{a_2x}{(x-a_1)(x-a_2)}\\~~~~~~~~~~+\frac{a_3x^2}{(x-a_1)(x-a_2)(x-a_3)} \\ \text{or,}~~y=\frac{x-a_1+a_1}{x-a_1}+\frac{a_2x}{(x-a_1)(x-a_2)} \\~~~~~~~~~~+ \frac{a_3x^2}{(x-a_1)(x-a_2)(x-a_3)} \\ \text{or,}~~y=\left[\frac{x}{x-a_1}+\frac{a_2x}{(x-a_1)(x-a_2)} \right] \\~~~~~~~~~~+ \frac{a_3x^2}{(x-a_1)(x-a_2)(x-a_3)}\\ \text{or,}~~ y=\frac{x(x-a_2)+a_2x}{(x-a_1)(x-a_2)} +\frac{a_3x^2}{(x-a_1)(x-a_2)(x-a_3)}\\ \text{or,}~~y=\frac{x^2}{(x-a_1)(x-a_2)}+\frac{a_3x^2}{(x-a_1)(x-a_2)(x-a_3)} \\ \text{or,}~~y=\frac{x^2(x-a_3)+a_3x^2}{(x-a_1)(x-a_2)(x-a_3)} \\ \text{or,}~~y=\frac{x^3}{(x-a_1)(x-a_2)(x-a_3)}\\ \therefore \log y=\log \left[\frac{x^3}{(x-a_1)(x-a_2)(x-a_3)}\right] \\ \text{or,}~~  \log y=\log x^3 \\~~~~~~~~~~-\log[(x-a_1)(x-a_2)(x-a_3)] \\ \text{or,}~~ \log y=3 \log x-\log(x-a_1)\\~~~~~~~~~~-\log(x-a_2)-\log(x-a_3) \\ \therefore \frac 1y~\frac{dy}{dx}=\frac 3x-\frac{1}{x-a_1}-\frac{1}{x-a_2}-\frac{1}{x-a_3}  \\ \text{or,}~~ \frac 1y~\frac{dy}{dx}=\left(\frac 1x-\frac{1}{x-a_1}\right) + \left(\frac 1x-\frac{1}{x-a_2}\right)\\~~~~~~~~~~ +\left(\frac 1x-\frac{1}{x-a_3}\right)  \\ \text{or,}~~ \frac 1y~\frac{dy}{dx}=\frac{x-a_1-x}{x(x-a_1)}\\~~~~~~~~~~+\frac{x-a_2-x}{x(x-a_2)}+\frac{x-a_3-x}{x(x-a_3)} \\ \text{or,}~~ \frac 1y~\frac{dy}{dx}=\frac{a_1}{x(a_1-x)}+\frac{a_2}{x(a_2-x)}+\frac{a_3}{x(a_3-x)} \\ \text{or,}~~  \frac{dy}{dx}=\frac yx\left[\frac{a_1}{a_1-x}+\frac{a_2}{a_2-x}+\frac{a_3}{a_3-x}\right]~~\text{(showed)}$

$~22.~$ If $~\frac{d}{dx} \left[\frac{\sin x}{\sin\left(x-\pi/4\right)}\right]=k \csc^2 (x-\pi/4),~~$ find $\,k\,$ and show that it is independent of $~x.$

Sol. $~~\frac{d}{dx} \left[\frac{\sin x}{\sin\left(x-\pi/4\right)}\right]\\=\frac{d}{dx} \left(\frac{\sin x}{\frac{1}{\sqrt 2} \sin x-\frac{1}{\sqrt 2}\cos x}\right)\\=\frac{d}{dx}\left(\frac{\sqrt 2 \sin x}{\sin x-\cos x}\right)\\=\sqrt 2\left[\frac{(\sin x-\cos x)\cos x-\sin x(\cos x+\sin x)}{(\sin x-\cos x)^2}\right]\\=\sqrt 2 \left[\frac{-(\cos^2x+\sin^2x)}{(\sin x-\cos x)^2}\right]\\=\sqrt 2 \left[\frac{-1}{2\sin^2\left(x-\frac{\pi}{4}\right)}\right]\\=-\frac{1}{\sqrt 2} \cdot \csc^2\left(x-\frac{\pi}{4}\right) \\ \therefore~ k \csc^2(x-\pi/4)=-\frac{1}{\sqrt 2}\csc^2(x-\pi/4) \\ \therefore k=-\frac{1}{\sqrt 2}\rightarrow(1)$

Hence, from $\,(1)\,$ we can conclude that it is independent of $~x.$