Question

Differentiate the following functions with respect to $x$
$(x\cos x{)}^x+(x\sin x{)}^{1/x}$

Answer 1

Let $y=(x\cos x{)}^x+(x\sin x{)}^{1/x}$. Then,

$y=e^{\log (x\cos x{)}^x}+e^{\log (x\sin x{)}^{1/x}}$ [ $\because y=a^{x}canbewrittenasy=e^{x\log a}$]

$\Rightarrow y=e^{x\log \left(x\cos x\right)}+e^{1/x\log \left(x\sin x\right)}$

$\Rightarrow \frac{d}{dx}\left\{e^{x\log \left(x\cos x\right)}\right\}+\frac{d}{dx}\left\{e^{1/x}\log \right(x\sin x\left)\right\}$

$\Rightarrow \frac{d}{dx}=e^{x\log \left(x\cos x\right)}\frac{d}{dx}|x\log \left(x\cos x\right)|+e^{1/x\log \left(x\sin x\right)}\frac{d}{dx}\left\{\frac{1}{x}\log \right(x\sin x\left)\right\}$

$\Rightarrow \frac{dy}{dx}=\left(x\cos x{)}^x\frac{d}{dx}|x\right(\log x+\log \cos x)|+(x\sin x{)}^{1/x}\frac{d}{dx}\left\{\frac{1}{x}\right(\log x+\log \sin x\left)\right\}$

$\Rightarrow \frac{dy}{dx}=(x\cos x{)}^x\left\{\right(\log x+\log \cos x)+x(\frac{1}{x}-\tan x)\}+(x\sin x{)}^{1/x}\{\frac{1}{x}(\frac{1}{x}+\cot x)-\frac{1}{x^2}(\log x+\log \sin x\left)\right\}$

$\frac{dy}{dx}=\left(x\cos x{)}^x|\log \right(x\cos x)+(1+x\tan x)|+(x\sin x{)}^{1/x}\{\frac{1}{x^2}+\frac{\cot x}{x}-\frac{\log x}{x^2}-\frac{\log \sin x}{x^2}\}$

$\frac{dy}{dx}=\left(x\sin x{)}^x|\log \right(x\cos x)+1-x\tan x|+(x\sin x{)}^{1/x}\left\{\frac{1+x\cot x-\log \left(x\sin x\right)}{x^2}\right\}$