Question

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

Answer 1

Let $y=x^{x\cos x}+\frac{x^2+1}{x^2-1}$
Also, let $u=x^{x\cos x}$ and $v=\frac{x^2+1}{x^2-1}$
$\therefore y=u+v$
$\Rightarrow \frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$ ...(1)
$u=x^{x\cos x}$
$\Rightarrow \log u=\log \left(x^{x\cos x}\right)$
$\Rightarrow \log u=x\cos x\log x$
Differentiating both sides with respect to $x$, we obtain
$\frac{1}{u}\frac{du}{dx}=\frac{d}{dx}\left(x\right).\cos x.\log x+x\frac{d}{dx}\left(\cos x\right).\log x+x\cos x.\frac{d}{dx}\left(\log x\right)$
$\Rightarrow \frac{du}{dx}=u[1.\cos x.\log x+x.(-\sin x)\log x+x\cos x.\frac{1}{x}]$
$\Rightarrow \frac{du}{dx}=x^{x\cos x}(\cos x.\log x-x\sin x\log x+\cos x)$
$\Rightarrow \frac{du}{dx}=x^{x\cos x}\left[\cos x\right(1+\log x)-x\sin x\log x]$ ...(2)
$v=\frac{x^2+1}{x^2-1}$
$\Rightarrow \log v=\log (x^2+1)-\log (x^2-1)$
Differentiating both sides with respect to $x$, we obtain
$\frac{1}{v}\frac{dv}{dx}=\frac{2x}{x^2+1}-\frac{2x}{x^2-1}$
$\Rightarrow \frac{dv}{dx}=v\left[\frac{2x(x^2-1)-2x(x^2+1)}{(x^2+1)(x^2-1)}\right]$
$\Rightarrow \frac{dv}{dx}=\frac{x^2+1}{x^2-1}\times \left[\frac{-4x}{(x^2+1)(x^2-1)}\right]$
$\Rightarrow \frac{dv}{dx}=\frac{-4x}{(x^2-1{)}^2}$ ...(3)
From (1), (2) and (3), we obtain
$\Rightarrow \frac{dy}{dx}=x^{x\cos x}\left[\cos x\right(1+\log x)-x\sin x\log x]-\frac{4x}{(x^2-1{)}^2}$