Question

Differentiate the given functions w.r.t. x.

$x^{xcosx}+\frac{x^2+1}{x^2-1}.$

Answer 1

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

Let $u=x^{xcosx},v=\frac{x^2+1}{x^2-1}$

y = u + v

Differentiating w.r.t. x,

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

Now, $u=x^{xcosx}$

Taking log on both sides, log u = x cos x log x

Differentiating w.r.t. x,

$\frac{d}{dx}\left(logu\right)=xcosx\frac{d}{dx}\left(logx\right)+logx\frac{d}{dx}\left(xcosx\right)=xcosx\times \frac{1}{x}+logx[x\frac{d}{dx}cosx+cosx\frac{d}{dx}(x\left)\right]\Rightarrow \frac{1}{u}\frac{du}{dx}=xcosx\times \frac{1}{x}=logx\left[\right(-xsinx)+cosx]\Rightarrow \frac{du}{dx}=u[cosx-xsinxlogx+cosx.logx]\Rightarrow \frac{du}{dx}=x^{xcosx}[cosx-xsinxlogx+cosx.logx]Again,v=\frac{x^2+1}{x^2-1}Takinglogonbothsides,logv=log(x^2+1)-log(x^2-1)Differentiatingw.r.t.x,\Rightarrow \frac{d}{dx}logv=\frac{d}{dx}log(x^2+1)-\frac{d}{dx}log(x^2-1)\Rightarrow \frac{1}{v}\frac{dv}{dx}=\frac{1}{(x^2+1)}\frac{d}{dx}(x^2+1)-\frac{1}{(x^2-1)}\frac{d}{dx}(x^2-1)\Rightarrow \frac{1}{v}\frac{dv}{dx}=\frac{2x}{(x^2+1)}-\frac{2x}{(x^2-1)}=2x\left[\frac{(x^2-1)-(x^2+1)}{\left(x_1^2\right)(x^2-1)}\right]\Rightarrow \frac{dv}{dx}=v\left[\frac{-4x}{x^4-1}\right]=\frac{x^2+1}{x^2-1}\left[\frac{-4x}{x^4-1}\right]Now,puttingthevaluesof\frac{du}{dx}and\frac{dv}{dx}inEq.\left(i\right)\therefore \frac{dy}{dx}=x^{xcosx}[cosx-xsinxlogx+cosx.logx]+\frac{x^2+1}{x^2-1}\left[\frac{-4x}{x^4-1}\right]=x^{xcosx}[cosx.(1+logx)-xsinxlogx]-\frac{4x}{(x^2-1{)}^2}$