Question

If $y={\left(\log x\right)}^{\cos x}+\frac{x^2+1}{x^2-1},$ find $\frac{dy}{dx}$

Answer 1

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

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

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

We have $u={\left(\log x\right)}^{\cos x}$ and $v=\frac{x^2+1}{x^2-1}$

$\Rightarrow \log u=\cos x\log x$ and $v=\frac{x^2+1}{x^2-1}$

$\Rightarrow \frac{1}{u}\frac{du}{dx}=\frac{\cos x}{x}-\log x\sin x$ and $\frac{dv}{dx}=\frac{(x^2-1)\frac{d}{dx}(x^2+1)-(x^2+1)\frac{d}{dx}(x^2-1)}{{(x^2-1)}^2}$

$\Rightarrow \frac{du}{dx}=u[\frac{\cos x}{x}-\log x\sin x]$ and $\frac{dv}{dx}=\frac{(x^2-1)\left(2x\right)-(x^2+1)\left(2x\right)}{{(x^2-1)}^2}$

$\Rightarrow \frac{du}{dx}={\left(\log x\right)}^{\cos x}[\frac{\cos x}{x}-\log x\sin x]$ and $\frac{dv}{dx}=\frac{2x(x^2-1-x^2-1)}{{(x^2-1)}^2}$

$\Rightarrow \frac{du}{dx}={\left(\log x\right)}^{\cos x}[\frac{\cos x}{x}-\log x\sin x]$ and $\frac{dv}{dx}=\frac{-4x}{{(x^2-1)}^2}$

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

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