Question

Find $\frac{dy}{dx}$, when $y=x^{x\cos x}+\left(\frac{x^2+1}{x^2-1}\right)$.

Answer 1

Given, $y=x^{x\cos x}+\left(\frac{x^2+1}{x^2-1}\right)$

Let $u=x^{x\cos x}$ and $v=\left(\frac{x^2+1}{x^2-1}\right)$. Then,

Now, $u=x^{x\cos x}$

Taking log on both side

$\log u=\left(x\cos x\right)\log x$

Differentiate with respect to $x$.

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

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

$\Rightarrow \frac{du}{dx}=x^{x\cos x}.\{\cos x-x\sin x(\log x)+\cos x(\log x\left)\right\}$

Now,$v=\left(\frac{x^2+1}{x^2-1}\right)$.

Differentiate with respect to $x$.

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

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

As $y=u+v$

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

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