Differentiation (Part-38) | S N De

$~(vi)~~(\sin x)^{\cos x}+(\cos x)^{\sin x}$

Sol. let $~~y=(\sin x)^{\cos x}+(\cos x)^{\sin x}=u+v~\text{(say)}\\ \therefore \frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}\rightarrow(1)$

Now, $~u=(\sin x)^{\cos x}\\ \text{or,}~~ \log u=\log(\sin x)^{\cos x} \\ \text{or,}~~ \log u=\cos x\log \sin x\\ \therefore \frac 1u \frac{du}{dx}=-\sin x\log(\sin x)\\~~~+\cos x \cdot \frac{1}{\sin x}\cdot \cos x\\ \therefore \frac 1u \frac{du}{dx}=-\sin x\log(\sin x)+\cos x \cot x \\ \text{or,}~~ \frac{du}{dx}=(\sin x)^{\cos x}[-\sin x\log(\sin x)\\~~~+\cos x \cot x]\rightarrow(2) $

Also, $~v=(\cos x)^{\sin x}\\ \text{or,}~~ \log v=\log (\cos x)^{\sin x} \\ \text{or,}~~ \log v=\sin x\log(\cos x)\\ \therefore \frac 1v \frac{dv}{dx}=\cos x\log(\cos x)\\~~~+\sin x \cdot \frac{1}{\cos x}\cdot (-\sin x) \\ \therefore \frac 1v \frac{dv}{dx}=\cos x\log(\cos x)-\sin x \tan x\\ \text{or,}~~ \frac{dv}{dx}=(\cos x)^{\sin x}[\cos x\log(\cos x)\\~~~-\sin x\tan x]\rightarrow(3)$

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

$~~\frac{dy}{dx}=(\sin x)^{\cos x}[-\sin x\log(\sin x)\\~~~+\cos x \cot x]\\~~~~+(\cos x)^{\sin x}[\cos x\log(\cos x)\\~~-\sin x\tan x]~~\text{(ans.)}$

$~(vii)~~e^{\cos^{-1}x}+x^{\sqrt x}$

Sol. $~~~y=e^{\cos^{-1}x}+x^{\sqrt x} \\ \text{or,}~~y=u+v~\text{(say)}\\ \text{or,}~~  \frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}\rightarrow(1)$

Now, $~u=e^{\cos^{-1}x} \\ \therefore \frac{du}{dx}=-\frac{e^{\cos^{-1}x}}{\sqrt{1-x^2}}\rightarrow(2)$

Again, $~~v=x^{\sqrt x} \\ \text{or,}~~  \log v=\log x^{\sqrt x} \\ \text{or,}~~  \log v=\sqrt x \log x \\ \therefore \frac 1v \frac{dv}{dx}=\frac{1}{2\sqrt x}\log x+\sqrt x \cdot \frac 1x \\ \text{or,}~~  \frac{dv}{dx}=x^{\sqrt x}\left(\frac{\log x}{2\sqrt x}+\frac{1}{\sqrt x}\right) \\ \text{or,}~~  \frac{dv}{dx}=x^{\sqrt x}\left(\frac{\log x+2}{2\sqrt x}\right)\rightarrow(3)$

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

$~\frac{dy}{dx}=-\frac{e^{\cos^{-1}x}}{\sqrt{1-x^2}}+x^{\sqrt x}\left(\frac{\log x+2}{2\sqrt x}\right)~~\text{(ans.)}$

$~(viii)~~(\sin x)^{\tan x}+(\cos x)^{\sec x}$

Sol. $~~y=(\sin x)^{\tan x}+(\cos x)^{\sec x} \\ \text{or,}~~  y=u+v~\text{(say)}\rightarrow(1)$

Now, $~u=(\sin x)^{\tan x} \\ \text{or,}~~  \log u=\log (\sin x)^{\tan x} \\ \text{or,}~~  \log u=\tan x \log(\sin x)\\ \therefore \frac 1u \frac{du}{dx}=\sec^2x \log(\sin x)\\~~+\tan x \cdot \frac{1}{\sin x}\cdot \cos x \\ \text{or,}~~  \frac{du}{dx}=(\sin x)^{\tan x}\\~~~\times [\sec^2x \log(\sin x)+1]\rightarrow(2)\\~~[\because~\tan x \cdot \frac{1}{\sin x}\cdot \cos x=\tan x \cdot \cot x=1]$

$~v=(\cos x)^{\sec x} \\ \text{or,}~~  \log v=\log (\cos x)^{\sec x} \\ \text{or,}~~  \log v=\sec x \log(\cos x) \\ \therefore \frac 1v \frac{dv}{dx}=\sec x\tan x\log(\cos x)\\~~~+\sec x \cdot \frac{1}{\cos x}\cdot (-\sin x) \\ \text{or,}~~  \frac{dv}{dx}=(\cos x)^{\sec x}[\sec x\tan x \log(\cos x)\\~~~-\sec x\tan x]\rightarrow(3) $

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

$~~\frac{dy}{dx}=(\sin x)^{\tan x}[\sec^2x \log(\sin x)+1]\\~~+(\cos x)^{\sec x}[\sec x\tan x \log(\cos x)\\~~~~-\sec x\tan x]~~\text{(ans.)}$

$~(ix)~~x^{\sin x}+(\sin x)^{\cos x}$

Sol. let $~~y=x^{\sin x}+(\sin x)^{\cos x}\\ \text{or,}~~y=u+v~~\text{(say)}\\ \therefore \frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}\rightarrow(1)$

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

Again, $~~v=(\sin x)^{\cos x} \\ \text{or,}~~ \log v=\log[(\sin x)^{\cos x}] \\ \text{or,}~~  \log v=\cos x \log(\sin x) \\ \therefore \frac 1v\frac{dv}{dx}=-\sin x\log(\sin x)\\~~~~+\cos x \cdot \frac{1}{\sin x}\cdot \cos x \\ \text{or,}~~  \frac{dv}{dx}=(\sin x)^{\cos x}\left[-\sin x\log(\sin x)\\~~~~+\cos x \cot x\right]\rightarrow(3)$

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

$~~\frac{dy}{dx}= x^{\sin x}\left(\cos x \log x+\frac{\sin x}{x}\right) \\~~~~~~~~+(\sin x)^{\cos x}\left[-\sin x\log(\sin x)\\~~~~+\cos x \cot x\right]~~\text{(ans.)}$

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

Sol. $~~~y^x+x^y+x^x=a^b$

Let $~~u=y^x,~~v=x^y,~w=x^x\\ ~~\text{so that}~~~~u+v+w=a^b \rightarrow(1)$

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

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

$~~~~w=x^x \\ \text{or,}~~\log w=\log x^x \\ \text{or,}~~ \log w=x \log x \\ \therefore \frac 1w \frac{dw}{dx}=1 \cdot \log x+ x \cdot \frac 1x \\ \text{or,}~~ \frac{dw}{dx}=x^x(\log x+1) $

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

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