Differentiation (Part-38) | S N De

 

$~14.~~$ If $~~y=\frac{x}{\sqrt{1+x^2}},~~$ prove that, $~x^3~\frac{dy}{dx}=y^3.$

Sol. $~~~~y=\frac{x}{\sqrt{1+x^2}} \\ \text{or,}~~y^2=\frac{x^2}{1+x^2}  \\ \text{or,}~~ \frac{1}{y^2}=\frac{1+x^2}{x^2}  \\ \text{or,}~~ \frac{1}{y^2}=\frac{1}{x^2}+1 \\ \therefore~ \frac{d}{dx}(y^{-2})=\frac{d}{dx}\left(\frac{1}{x^2}+1\right)  \\ \text{or,}~~ -2y^{-3}~\frac{dy}{dx}\\~~~~~=\frac{d}{dx}(x^{-2})+0 \\~~~~~=-2x^{-3}  \\ \text{or,}~~ y^{-3}~\frac{dy}{dx}=x^{-3}  \\ \text{or,}~~ x^3~\frac{dy}{dx}=y^3~~\text{(proved)}$

$~15(i)~$ If $~~y=e^{y/x}~~$ show that, $~~\frac{dy}{dx}=\frac{y^2}{x(y-x)}.$

Sol. $~~~~~y=e^{y/x} \\ \therefore \frac{dy}{dx}\\=\frac{d}{dx}\left(e^{y/x}\right)\\=e^{y/x}~\frac{d}{dx}(y/x)\\=e^{y/x}\left[y~\frac{d}{dx}(\frac 1x)+\frac 1x~\frac{dy}{dx}\right]\\=y\left[y~\left(-\frac{1}{x^2}\right)+\frac 1x~\frac{dy}{dx}\right]\\=-\frac{y^2}{x^2}+\frac yx~\frac{dy}{dx} \\ \text{or,}~~\frac{dy}{dx}=\frac yx~\frac{dy}{dx}-\frac{y^2}{x^2} \\ \text{or,}~~\left(1-\frac yx\right)~\frac{dy}{dx}=-\frac{y^2}{x^2} \\ \text{or,}~~x^2\left(1-\frac yx\right)~\frac{dy}{dx}=-y^2 \\ \text{or,}~~(x^2-xy)~\frac{dy}{dx}=-y^2 \\ \therefore \frac{dy}{dx}=\frac{-y^2}{-x(y-x)}=\frac{y^2}{x(y-x)}~~\text{(showed)}$

$~~15(ii)~~$ If $~ye^y=x,~~$ prove that, $~~\frac{dy}{dx}=\frac{y}{x(1+y)}.$

Sol. $~~~~ye^y=x \rightarrow(1) \\ \therefore \frac{d}{dx}(ye^y)=\frac{d}{dx}(x) \\ \text{or,}~~\frac{dy}{dx}~e^y+y~\frac{d}{dx}(e^y)=1 \\ \text{or,}~~ \frac{dy}{dx}~\frac xy+ye^y~\frac{dy}{dx}=1~~[\text{By (1)}] \\ \text{or,}~~ \frac{dy}{dx}~\frac xy+x~\frac{dy}{dx}=1 \\ \text{or,}~~ \left(\frac xy+x\right)~\frac{dy}{dx}=1 \\ \text{or,}~~ x\left(\frac 1y+1\right)~\frac{dy}{dx}=1 \\ \text{or,}~~\left(\frac{1+y}{y}\right)~\frac{dy}{dx}=\frac 1x \\ \text{or,}~~ \frac{dy}{dx}=\frac{y}{x(1+y)}~~\text{(proved)}$

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

Sol.  let $~~x=\sin\theta,~~y=\sin\phi.$

Now, $~~\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)\\ \Rightarrow \sqrt{1-\sin^2\theta}+\sqrt{1-\sin^2\phi}\\~~~~~=a(\sin\theta-\sin\phi)\\ \Rightarrow \cos\theta+\cos\phi=a(\sin\theta-\sin\phi)\\ \Rightarrow \frac{2\cos\left(\frac{\theta+\phi}{2}\right)\cos\left(\frac{\theta-\phi}{2}\right)}{2\cos\left(\frac{\theta+\phi}{2}\right)\sin\left(\frac{\theta-\phi}{2}\right)}=a \\ \therefore~~\cot\left(\frac{\theta-\phi}{2}\right)=a \\ \text{or,}~~\frac{\theta-\phi}{2}=\cot^{-1}a \\ \text{or,}~~ \theta-\phi=2\cot^{-1}a \\ \text{or,}~~ \sin^{-1}x-\sin^{-1}y=2\cot^{-1}a \\ \therefore ~\frac{d}{dx}(\sin^{-1}x-\sin^{-1}y)=\frac{d}{dx}(2\cot^{-1}a) \\ \text{or,}~~ \frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-y^2}}\cdot \frac{dy}{dx}=0 \\ \text{or,}~~ \frac{dy}{dx}=\frac{\sqrt{1-y^2}}{\sqrt{1-x^2}}~~\text{(showed)}$

$~16.~$ If $~~y=e^{\sin^{-1}x}~$ and $~~z=e^{-\cos^{-1}x},~~$ prove that the value of $~~\frac{dy}{dz}~$ is independent of $~x.$

Sol. From the given condition, we have,

$~~~~~~~\frac yz=\frac{e^{\sin^{-1}x}}{e^{-\cos^{-1}x}}=e^{\sin^{-1}x+\cos^{-1}x} \\ \therefore ~~\frac yz=e^{\pi/2} \\ \text{or,}~~y=ze^{\pi/2}\\ \text{or,}~~\frac{d}{dz}(y)=\frac{d}{dz}(ze^{\pi/2})\\ \text{or,}~~ \frac{dy}{dz}=e^{\pi/2} \rightarrow(1)$

Hence, from $~(1),~$ we can conclude that  the value of $~~\frac{dy}{dz}~$ is independent of $~x.$