Question

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

Answer 1

Let,

$y_1=x^{x\cos x}.............\left(1\right)\log y_1=\log \left(x^{x\cos x}\right)\log y_1=x\cos x\log \left(x\right)\frac{d{y}_1}{dx}\frac{1}{y_1}=xcosx\frac{1}{x}+\log x\left(\frac{d}{dx}\right(x\cos x\left)\right)=x\cos x\frac{1}{x}+\log x(-x\sin x+\cos x)=\cos x+\log x(-x\sin x+\cos x)=\cos x-x\log x\sin x+\log x\cos x\frac{d{y}_1}{dx}=y_1(\cos x-x\log x\sin x+\log x\cos x)\frac{d{y}_1}{dx}=\left(x^{x\cos x}\right)(cosx-x\log x\sin x+\log x\cos x).................\left(2\right)$

Now, from (1)

Let $y=x^{x\cos x}+\frac{x^2+1}{x^2-1}\Rightarrow y=y_1+\frac{x^2+1}{x^2-1}\Rightarrow y=y_1+\frac{x^2-1+2}{x^2-1}\Rightarrow y=y_1+1+\frac{2}{x^2-1}\frac{dy}{dx}=\frac{d{y}_1}{dx}+0-2(x^2-1{)}^{-2}.2x...................(3)$

So, From (2) and (3)

$\frac{dy}{dx}=\left(x^{x\cos x}\right)(\cos x-x\log x\sin x+\log x\cos x)-(x^2-1{)}^{-2}.4x$