Question

Differentiate
$x^{x\cos x}+\frac{x^2+1}{x^2-1}$ w.r.t.x.

Answer 1

Let $y=x^{x\cos x}+\frac{x^2+1}{x^2-1}$

$\Rightarrow y=u+v$

$\therefore u=x^{x\cos x}$

$\Rightarrow \log u=x\cos x\log x$

$\Rightarrow \frac{1}{u}\frac{du}{dx}=x\cos x\frac{d}{dx}\left(\log x\right)+x\log x\frac{d}{dx}\left(\cos x\right)+\cos x\log x\frac{d}{dx}\left(x\right)$

$=x\cos x\times \frac{1}{x}-x\log x\sin x+\cos x\log x$

$\frac{1}{u}\frac{du}{dx}=\cos x-x\log x\sin x+\cos x\log x$

$\frac{du}{dx}=u(\cos x-x\log x\sin x+\cos x\log x)$

$\therefore \frac{du}{dx}=x^{x\cos x}(\cos x-x\log x\sin x+\cos x\log x)$

$v=\frac{x^2+1}{x^2-1}$

$\frac{dv}{dx}=\frac{(x^2-1)\frac{d}{dx}(x^2+1)-(x^2+1)\frac{d}{dx}(x^2-1)}{{(x^2-1)}^2}$

$\frac{dv}{dx}=\frac{(x^2-1)\left(2x\right)-(x^2+1)\left(2x\right)}{{(x^2-1)}^2}$

$\frac{dv}{dx}=\frac{2x(x^2-1-x^2-1)}{{(x^2-1)}^2}$

$\frac{dv}{dx}=\frac{-4x}{{(x^2-1)}^2}$

$\therefore \frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$

$=x^{x\cos x}(\cos x-x\log x\sin x+\cos x\log x)-\frac{4x}{{(x^2-1)}^2}$