Differentiation (Part-38) | S N De

Differentiation (Part-19) | S N De

$~17.~~$ If $~~y=\sin\left(\frac{\pi}{6}e^{xy}\right),~$ find $~~\frac{dy}{dx}~$ at $~x=0.$

Sol.  $~y=\sin\left(\frac{\pi}{6}e^{xy}\right)\rightarrow(1) \\ \therefore \frac{dy}{dx}=\frac{d}{dx}\left[\sin \left(\frac{\pi}{6}e^{xy}\right)\right]\\ \text{or,}~~\frac{dy}{dx}\\=\cos\left(\frac{\pi}{6}e^{xy}\right)~\cdot \frac{d}{dx}\left(\frac{\pi}{6}e^{xy}\right)\\=\cos\left(\frac{\pi}{6}e^{xy}\right)\cdot \frac{\pi}{6}e^{xy}\frac{d}{dx}(xy)\\=\cos\left(\frac{\pi}{6}e^{xy}\right)\cdot \frac{\pi}{6}e^{xy}\left(x~\frac{dy}{dx}+y\right)\\=\left(\frac{\pi}{6}xe^{xy}\right)\cos\left(\frac{\pi}{6}e^{xy}\right)~\frac{dy}{dx}\\~~~~~+\frac{\pi}{6}ye^{xy}\cos\left(\frac{\pi}{6}e^{xy}\right)\\ \text{or,}~~\left[1-\frac{\pi}{6}xe^{xy}\cos\left(\frac{\pi}{6}e^{xy}\right)\right]~\frac{dy}{dx}\\~~~~=\frac{\pi}{6}ye^{xy}\cos\left(\frac{\pi}{6}e^{xy}\right) \\ \text{or,}~~\frac{dy}{dx}=\frac{(\pi/6)ye^{xy}\cos\left(\frac{\pi}{6}e^{xy}\right)}{1-(\pi/6)xe^{xy}\cos\left(\frac{\pi}{6}e^{xy}\right)}$

Now, from $\,(1)\,$ we get, for $~x=0,~$

$~~y=\sin\left(\frac{\pi}{6}\right)=\frac 12\\ \therefore~~\left[\frac{dy}{dx}\right]_{x=0}\\=\frac{(\pi/6)\cdot \frac 12e^{0}\cdot \cos\left(\frac{\pi}{6}e^{0}\right)}{1-0}\\=\frac{\pi}{12} \cdot \frac{\sqrt 3}{2}\\=\frac{\sqrt 3 \pi}{24}~~\text{(ans.)}$

$~18(i)~~$ If $~~y=x^{x^{x^{\cdots \infty}}},~~$ then prove that, $~~\frac{dy}{dx}=\frac{y^2}{x(1-y\log x)}.$

Sol. $~~~~~~y=x^{x^{x^{\cdots \infty}}} \\\Rightarrow y=x^y \\ \therefore~~\log(y)=\log(x^y)=y\log x \\ \text{or,}~~ \frac{d}{dx}(\log y)=\frac{d}{dx}(y\log x) \\ \text{or,}~~ \frac 1y\cdot \frac{dy}{dx}=y~\cdot \frac 1x+\log x ~\cdot \frac{dy}{dx}\\ \text{or,}~~ \left(\frac 1y-\log x\right)~\frac{dy}{dx}=\frac yx \\ \text{or,}~~ \frac{dy}{dx}=\frac{y^2}{x(1-y\log x)}~~\text{(proved)} $

$~18(ii)~~$ If $~~y=\sqrt x^{\sqrt x^{\sqrt x^{\cdots \infty}}},~~$ then prove that, $~~x\frac{dy}{dx}=\frac{y^2}{(1-y\log x)}.$

Sol. $~~~~~~y=\sqrt x^{\sqrt x^{\sqrt x^{\cdots \infty}}} \\\Rightarrow y=(\sqrt x)^y \\ \therefore~~\log(y)=\log(\sqrt x)^y=y\log x^{1/2} \\ \text{or,}~~ \frac{d}{dx}(\log y)=\frac 12~\frac{d}{dx}(y\log x) \\ \text{or,}~~ \frac 2y\cdot \frac{dy}{dx}=y~\cdot \frac 1x+\log x ~\cdot \frac{dy}{dx}\\ \text{or,}~~ \left(\frac 2y-\log x\right)~\frac{dy}{dx}=\frac yx \\ \text{or,}~~ (2-y\log x)~\frac{dy}{dx}=\frac{y^2}{x} \\ \text{or,}~~ x~\frac{dy}{dx}=\frac{y^2}{2-y\log x}~~\text{(proved)} $

$~(iii)~$ If $~~y=x+\frac{1}{x+\frac{1}{x+\cdots \infty}},~~$ prove that, $~~\frac{dy}{dx}=\frac{y}{2y-x}.$

Sol. $~~~~~~~y=x+\frac{1}{x+\frac{1}{x+\cdots \infty}} \\ \text{or,}~~ y=x+\frac 1y \\ \text{or,}~~ y^2=xy+1 \\ \therefore~\frac{d}{dx}(y^2)=\frac{d}{dx}(xy+1) \\ \text{or,}~~ 2y~\frac{dy}{dx}=x~\frac{dy}{dx}+y \cdot 1+0 \\ \text{or,}~~ (2y-x)~\frac{dy}{dx}=y \\ \text{or,}~~ \frac{dy}{dx}=\frac{y}{2y-x}~~\text{(proved)}$

$~19.~~$ Show that the derivative of $~~\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}~~$ w.r.t. $~~\sqrt{1-x^4}~$ is $~~\frac{\sqrt{1-x^4}-1}{x^6}.$

Sol. let $~u=\sqrt{1-x^4} \rightarrow(1)$ and 

$~~v=\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}} \\ \text{or,}~~  v^2=\frac{1+x^2-2\sqrt{(1+x^2)(1-x^2)}~~+1-x^2}{1+x^2+2\sqrt{(1+x^2)(1-x^2)}+1-x^2} \\ \text{or,}~~  v^2=\frac{2-2\sqrt{1-(x^2)^2}}{2+2\sqrt{1-(x^2)^2}} \\ \text{or,}~~  v^2=\frac{2(1-\sqrt{1-x^4})}{2(1+\sqrt{1-x^4})} \\ \text{or,}~~  v^2=\frac{1-u}{1+u}~~~~[~\text{By (1)}~] \\ \text{or,}~~  v=\frac{\sqrt{1-u}}{\sqrt{1+u}} \\ \therefore ~\frac{dv}{du}\\=\frac{d}{du} \left(\frac{\sqrt{1-u}}{\sqrt{1+u}}\right)\\=\frac{-\frac{\sqrt{1+u}}{2\sqrt{1-u}}-\frac{\sqrt{1-u}}{2\sqrt{1+u}}}{(\sqrt{1+u})^2}\\=\frac{-1-u-1+u}{2\sqrt{(1-u)(1+u)}~(1+u)}\\=\frac{-2}{2\sqrt{1-u^2}~(1+u)}\\=\frac{-1}{\sqrt{1-1+x^4}~(1+\sqrt{1-x^4})}\\=\frac{-(1-\sqrt{1-x^4})}{x^2(1+\sqrt{1-x^4})(1-\sqrt{1-x^4})}\\=\frac{\sqrt{1-x^4}-1}{x^2(1-1+x^4)}~~[*]\\=\frac{\sqrt{1-x^4}-1}{x^6}~~\text{(ans.)}$

Note[*] : $~~~~(1+\sqrt{1-x^4})(1-\sqrt{1-x^4})\\=(1)^2-(\sqrt{1-x^4})^2\\=1-(1-x^4)\\=1-1+x^4\\=x^4$