Differentiation (Part-38) | S N De

$~4.~$ Find $\,\frac{dy}{dx}\,$ when:

$~(i)~~~x^y+y^x=a$

Sol. let $~~u=x^y,~~v=y^x ~$ so that $~~u+v=a~$ which gives $~~\frac{du}{dx}+\frac{dv}{dx}=0 \rightarrow(1)$

Now, $~~~~~u=x^y \\ \text{or,}~~\log u=\log (x^y)\\ \text{or,}~~  \log u=y\log x\\ \therefore \frac{1}{u} \frac{du}{dx}=\frac yx+\log x~\frac{dy}{dx} \\ \text{or,}~~  \frac{du}{dx}=x^y\left(\frac yx+\log x~\frac{dy}{dx}\right) \rightarrow(2)$

Again, $~~~~~v=y^x \\ \text{or,}~~\log v=\log (y^x)\\ \text{or,}~~  \log v=x\log y\\ \therefore \frac{1}{v} \frac{dv}{dx}=\frac xy~\frac{dy}{dx}+1 \cdot \log y \\ \text{or,}~~  \frac{dv}{dx}=y^x\left(\frac xy~\frac{dy}{dx}+ \log y \right) \rightarrow(3)$

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

$~~~x^y\left(\frac yx+\log x~\frac{dy}{dx}\right) \\~~~~+y^x\left(\frac xy~\frac{dy}{dx}+ \log y \right) =0 \\ \text{or,}~~ x^{y-1} .y+x^y .\log x .\frac{dy}{dx}\\~~~+xy^{x-1} ~\frac{dy}{dx}+y^x. \log y=0 \\ \text{or,}~~  \frac{dy}{dx}(x^y \log x+xy^{x-1})\\~~~~=-(x^{y-1}.y+y^x \log y) \\ \text{or,}~~  \frac{dy}{dx}=-\frac{x^{y-1}.y+y^x \log y}{x^y \log x+xy^{x-1}}~~\text{(ans.)}$

$~(ii)~~y=\log \sqrt{\frac{1-\cos x}{1+\cos x}}+a^x$

Sol. $~~~y=\log \sqrt{\frac{1-\cos x}{1+\cos x}}+a^x \\ \text{or,}~~  y=\log \sqrt{\frac{2\sin^2(x/2)}{2\cos^2(x/2)}}+a^x \\ \text{or,}~~  y=\log \left(\frac{\sin(x/2)}{\cos(x/2)}\right)+a^x \\ \text{or,}~~ y=\log(\tan(x/2))+a^x \\ \therefore \frac{dy}{dx}=\frac{1}{\tan()x/2} \cdot \sec^2(x/2) \cdot \frac 12+\frac{d}{dx}(a^x) \\ \text{or,}~~ \frac{dy}{dx}=\frac{\cos(x/2)}{\sin(x/2)} \cdot \frac{1}{2\cos^2(x/2)}+a^x \log_ea \\ \text{or,}~~  \frac{dy}{dx}=\frac{1}{2\sin(x/2)\cos(x/2)}+a^x \log_ea \\ \therefore \frac{dy}{dx}=\frac{1}{\sin x}+a^x \log_ea \\ \text{or,}~~ \frac{dy}{dx}=\csc x+a^x \log_ea$

Note $\,1:~~\csc x \rightarrow \text{cosec }x$

Note $\,2:~~\text{let}~~w=a^x \\ \Rightarrow \log w=\log(a^x) \\ \Rightarrow \log_e w=x \log_ea \\ \therefore \frac 1w \frac{dw}{dx} =1 \cdot \log_ea \\ \Rightarrow \frac{dw}{dx}=a^x \cdot \log_e a$ 

$~(iii)~~y=\frac{x^3 \cdot \sqrt{x^2-12}}{\sqrt[3]{20-3x}}~~\text{when}~~x=4.$

Sol. $~~~y=\frac{x^3 \cdot \sqrt{x^2-12}}{3\sqrt{20-3x}} \\ \text{or,}~~ \log_e y=\log_e\left(\frac{x^3 \cdot \sqrt{x^2-12}}{\sqrt[3]{20-3x}}\right) \\ \text{or,}~~  \log y=\log(x^3\sqrt{x^2-12})\\~~~~~~~~~-\log(\sqrt[3]{20-3x}) \\ \text{or,}~~\log y=\log x^3+ \log(x^2-12)^{1/2}\\~~~~~~~~~~~-\log(20-3x)^{1/3} \\ \text{or,}~~ \log y=3\log x+\frac 12\log(x^2-12)\\~~~~~~~~~-\frac 13 \log(20-3x) \\ \therefore \frac 1y \frac{dy}{dx}=\frac 3x+\frac 12 \cdot \frac{1}{x^2-12} \cdot 2x\\~~~~~~~-\frac 13\cdot\frac{1}{20-3x} \cdot (-3) \\ \text{or,}~~ \frac 1y ~\frac{dy}{dx}=\frac 3x+\frac{x}{x^2-12}\\~~+\frac{1}{20-3x} \\ \text{or,}~~ ~\frac{dy}{dx}=y\left(\frac 3x+\frac{x}{x^2-12}+\frac{1}{20-3x}\right)\rightarrow(1)$

Now, the value of $~y~$ at $~x=4~$ is :

$=\frac{4^3\sqrt{4^2-12}}{\sqrt[3]{20-3 \times 4}}=\frac{64 \times 2}{\sqrt[3]{8}}=\frac{64 \times 2}{2}=64\rightarrow(2)$

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

$~~~~~\left[\frac{dy}{dx}\right]_{x=4}=64 \times  \left(\frac 34+\frac{4}{4^2-12}+\frac{1}{20-3 \times 4}\right) \\ \therefore ~~\left[\frac{dy}{dx}\right]_{x=4}=64 \times\left( \frac 34+1+\frac 18\right) \\~~~~~~~~~~~~~~~~~~~~~= 64 \cdot \left(\frac{6+8+1}{8}\right) \\~~~~~~~~~~~~~~~~~~~~~= 64 \times \frac{15}{8} \\~~~~~~~~~~~~~~~~~~~~~= 8 \times 15 \\~~~~~~~~~~~~~~~~~~~~~= 120~~\text{(ans.)}$

$~(iv)~~\tan^2y=\frac{1+\cos 2x}{1-\cos2x}$

Sol. $~~\tan^2y=\frac{1+\cos 2x}{1-\cos2x} \\ \text{or,}~~ \tan^2y=\frac{2\cos^2x}{2\sin^2x} \\ \text{or,}~~ \tan^2y=\cot^2x \\ \text{or,}~~ \tan y= \pm \cot x =\cot(\pm  x) \\ \text{or,}~~ \tan y=\tan\left( \pi/2-(\pm x)\right) \\ \text{or,}~~ y=\pi/2 \mp x \\ \therefore \frac{dy}{dx}=0 \mp \frac{d}{dx} (x) =\mp 1 \\ \therefore \frac{dy}{dx} =\pm1~~\text{(ans.)}$

$~(v)~~e^y-\frac{a+b\tan x}{a-b\tan x}=0$

Sol. $~~e^y-\frac{a+b\tan x}{a-b\tan x}=0 \\ \text{or,}~~ e^y=\frac{a+b\tan x}{a-b\tan x} \\ \therefore e^y~\frac{dy}{dx}=\frac{b\sec^2x(a-b\tan x)+b\sec^2x(a+b\tan x)}{(a-b\tan x)^2} \\ \text{or,}~~ \frac{dy}{dx}=\frac{1}{e^y} \cdot \frac{b\sec^2x(a-b\tan x+a+b\tan x)}{(a-b\tan x)^2} \\ \text{or,}~~ \frac{dy}{dx}=\frac{1}{e^y}\cdot \frac{2ab \sec^2x}{(a-b\tan x)^2} \\ \text{or,}~~ \frac{dy}{dx}=\frac{a-b\tan x}{a+b\tan x} \cdot \frac{2ab \sec^2x}{(a-b\tan x)^2} \\ \text{or,}~~ \frac{dy}{dx}=\frac{2ab\sec^2x}{(a+b\tan x)(a-b\tan x)} \\ \text{or,}~~ \frac{dy}{dx}=\frac{2ab \sec^2x}{a^2-b^2\tan^2x} \\ \text{or,}~~ \frac{dy}{dx}=\frac{2ab/(\cos^2x)}{a^2-b^2  \cdot \frac{\sin^2x}{\cos^2x}} \\ \text{or,}~~ \frac{dy}{dx}=\frac{2ab}{a^2\cos^2x-b^2\sin^2x}~~\text{(ans.)}$

$~(vi)~~y=2\tan^{-1}\sqrt{\frac{x-a}{b-x}}$

Sol. $~~y=2\tan^{-1}\sqrt{\frac{x-a}{b-x}} \\ \therefore \frac{dy}{dx}=\frac{2}{1+\left(\sqrt{\frac{x-a}{b-x}}\right)^2} \times \frac{d}{dx} \left(\sqrt{\frac{x-a}{b-x}}\right) \\ \text{or,}~~\frac{dy}{dx}=\frac{2}{1+\frac{x-a}{b-x}} \times  \frac{\sqrt{b-x} \cdot \frac{1}{2\sqrt{x-a}}-\sqrt{x-a} \left(\frac{-1}{\sqrt{b-x}}\right)}{(\sqrt{b-x})^2} \\ \text{or,}~~\frac{dy}{dx}=\frac{2(b-x)}{b-x+x-a} \times  \frac{1}{b-x} \\~~~~\times  \frac 12\left[\sqrt{\frac{b-x}{x-a}} +\sqrt{\frac{x-a}{b-x}} \right] \\ \text{or,}~~\frac{dy}{dx}=\frac{1}{b-a} \times  \frac{b-x+x-a}{\sqrt{(x-a)(b-x)}} \\~ \therefore  \frac{dy}{dx}= \frac{1}{b-a} \times  \frac{b-a}{\sqrt{(x-a)(b-x)}} \\ \text{or,}~~\frac{dy}{dx}=\frac{1}{\sqrt{(x-a)(b-x)}}~~\text{(ans.)}$