Question

Differentiate the function w.r.t. $x$.
$(x\cos x{)}^x+(x\sin x{)}^{\frac{1}{x}}$

Answer 1

Let $y=(x\cos x{)}^x+(x\sin x{)}^{\frac{1}{x}}$
Also, let $u=(x\cos x{)}^x$ and $v=(x\sin x{)}^{\frac{1}{x}}$
$\therefore y=u+v$
$\Rightarrow \frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$ ...(1)
$u=(x\cos x{)}^x$
$\Rightarrow \log u=\log (x\cos x{)}^x$
$\Rightarrow \log u=x\log \left(x\cos x\right)$
$\Rightarrow \log u=x[\log x+\log \cos x]$
$\Rightarrow \log u=x\log x+x\log \cos x$
Differentiating both sides with respect to $x$, we obtain
$\frac{1}{u}\frac{du}{dx}=\frac{d}{dx}\left(x\log x\right)+\frac{d}{dx}\left(x\log \cos x\right)$
$\Rightarrow \frac{du}{dx}=u[(\log x.\frac{d}{dx}(x)+x.\frac{d}{dx}(\log x\left)\right)+(\log \cos x.\frac{d}{dx}(x)+x.\frac{d}{dx}(\log \cos x\left)\right)]$
$\Rightarrow \frac{du}{dx}=(x\cos x{)}^x[(\log x.1+x.\frac{1}{x})+(\log \cos x.1+x.\frac{1}{\cos x}.\frac{d}{dx}(\cos x\left)\right)]$
$\Rightarrow \frac{du}{dx}=(x\cos x{)}^x[(\log x+1)+(\log \cos x+\frac{x}{\cos x}.(-\sin x\left)\right)]$
$\Rightarrow \frac{du}{dx}=(x\cos x{)}^x[(1+\log x)+(\log \cos x-x\tan x)]$
$\Rightarrow \frac{du}{dx}=(x\cos x{)}^x[1-x\tan x+(\log x+\log \cos x)]$
$\Rightarrow \frac{du}{dx}=(x\cos x{)}^x[1-x\tan x+\log (x\cos x\left)\right]$ ... (2)
$v=(x\sin x{)}^{\frac{1}{x}}$
$\Rightarrow \log v=\log (x\sin x{)}^{\frac{1}{x}}$
$\Rightarrow \log v=\frac{1}{x}\log \left(x\sin x\right)$
$\Rightarrow \log v=\frac{1}{x}(\log x+\log \sin x)$
$\Rightarrow \log v=\frac{1}{x}\log x+\frac{1}{x}\log \sin x$
Differentiating both sides with respect to $x$, we obtain
$\frac{1}{v}\frac{dv}{dx}=\frac{d}{dx}\left(\frac{1}{x}\log x\right)+\frac{d}{dx}\left[\frac{1}{x}\log \right(\sin x\left)\right]$
$\Rightarrow \frac{1}{v}\frac{dv}{dx}=[\log x.\frac{d}{dx}\left(\frac{1}{x}\right)+\frac{1}{x}.\frac{d}{dx}(\log x\left)\right]+\left[\log \right(\sin x).\frac{d}{dx}\left(\frac{1}{x}\right)+\frac{1}{x}.\frac{d}{dx}(\log \left(\sin x\right)\left)\right]$
$\Rightarrow \frac{1}{v}\frac{dv}{dx}=[\log x.(-\frac{1}{x^2})+\frac{1}{x}.\frac{1}{x}]+\left[\log \right(\sin x).(-\frac{1}{x^2})+\frac{1}{x}.\frac{1}{\sin x}.\frac{d}{dx}(\sin x\left)\right]$
$\Rightarrow \frac{1}{v}\frac{dv}{dx}=\frac{1}{x^2}(1-\log x)+[-\frac{\log \left(\sin x\right)}{x^2}+\frac{1}{x\sin x}.\cos x]$
$\Rightarrow \frac{dv}{dx}=(x\sin x{)}^{\frac{1}{x}}[\frac{1-\log x}{x^2}+\frac{-\log \left(\sin x\right)+x\cot x}{x^2}]$
$\Rightarrow \frac{dv}{dx}=(x\sin x{)}^{\frac{1}{x}}\left[\frac{1-\log x-\log \left(\sin x\right)+x\cot x}{x^2}\right]$
$\Rightarrow \frac{dv}{dx}=(x\sin x{)}^{\frac{1}{x}}\left[\frac{1-\log \left(x\sin x\right)+x\cot x}{x^2}\right]$ ...(3)
From (1), (2) and (3), we obtain
$\frac{dy}{dx}=(x\cos x{)}^x[1-x\tan x+\log (x\cos x\left)\right]+(x\sin x{)}^{\frac{1}{x}}\left[\frac{x\cot x+1-\log \left(x\sin x\right)}{x^2}\right]$