Question

If $y=\frac{x^{\sin x}}{1+x+x^2}$, Find the $\frac{dy}{dx}$

Answer 1

$y=\frac{x^{\sin x}}{1+x+x^2}$

$\Rightarrow (1+x+x^2)y=x^{\sin x}$

Let $u=x^{\sin x}$

$\Rightarrow \log u=\log x^{\sin x}$ by applying $\log$ to both sides.

$\Rightarrow \log u=\sin x\log x$ using $\log a^m=m\log a$

$\Rightarrow \frac{1}{u}\frac{du}{dx}=\frac{\sin x}{x}+\log x\cos x$

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

$\therefore \frac{d}{dx}\left(x^{\sin x}\right)=x^{\sin x}(\frac{\sin x}{x}+\cos x\log x)$

$\Rightarrow (1+x+x^2)\frac{dy}{dx}+(1+2x)y=\frac{d}{dx}\left(x^{\sin x}\right)$

$\Rightarrow (1+x+x^2)\frac{dy}{dx}=x^{\sin x}(\frac{\sin x}{x}+\cos x\log x)-(1+2x)y$

$\therefore \frac{dy}{dx}=\frac{\sin x{x}^{\sin x-1}+x^{\sin x}\cos x\log x-y-2xy}{1+x+x^2}$