Question

Differentiate the function with respect to $x$
$x^{xcos\,x}+\frac{x^2+1}{x^2-1}$

Answer 1

The correct answer is $\frac{dy}{dx}=x^{xcosx}[cosx(1+logx)-xsinx\,logx]-\frac{4x}{(x^2-1)^2}$
Let $y=x^{xcos\,x}+\frac{x^2+1}{x^2-1}$
Also,let $u=x^{xcos\,x}$ and $v=\frac{x^2+1}{x^2-1}$
$∴y=u+v$
$⇒\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx} ....(1)$
$u=x^{xcos\,x}$
$⇒log\,u=log(x^{xcos\,x})$
$⇒log\,u=xcos\,x\,logx$
Differentiating both sides with respect to $x$,we obtain
$⇒\frac{1}{u}\frac{du}{dx}=\frac{d}{dx}(x).cosx.log(x)+x.\frac{d}{dx}(cosx).logx+xcosx.\frac{d}{dx}[logx]$
$⇒\frac{du}{dx}=u[1.cosx.logx+x.(-sinx)logx+xcosx.\frac{1}{x}]$
$⇒\frac{du}{dx}=x^{xcosx}[cosx\,logx-xsinx\,logx+cosx]$
$⇒\frac{du}{dx}=x^{xcosx}[cosx(1+logx)-xsinx\,logx] ..(2)$
$v=\frac{x^2+1}{x^2-1}$
$⇒log\,v=log(x^2+1)-log(x^2-1)$
Differentiating both sides with respect to $x$, we obtain
$\frac{1}{v}\frac{dv}{dx}=\frac{2x}{x^2+1}-\frac{2x}{x^2-1}$
$⇒\frac{dv}{dx}=v[\frac{2x(x^2-1)-2x(x^2+1)}{(x^2+1)(x^2-1)}]$
$⇒\frac{dv}{dx}=\frac{x^2+1}{x^2-1}\times [\frac{-4x}{(x^2+1)(x^2-1)}]$
$⇒\frac{dv}{dx}=\frac{-4x}{(x^2-1)^2} ....(3)$
Therefore,from (1),(2),and(3),we obtain
$\frac{dy}{dx}=x^{xcosx}[cosx(1+logx)-xsinx\,logx]-\frac{4x}{(x^2-1)^2}$