Question

Differentiate w.r.t. x:
$y=x^{sinx}+\sqrt{x}$

Answer 1

$y=x^{\sin x}+\sqrt{x}$

Let $x^{\sin x}=2$

$y=z+\sqrt{x}$

Differentiating both sides

$\frac{dy}{dx}=\frac{dz}{dx}+\frac{1}{2\sqrt{x}}$ ....... $\left(1\right)$

$z=x^{\sin x}$ ............ $\left(2\right)$

Taking $\ln$ on both sides

$\ln z=\sin x\ln x$

Differentiating both sides

$\frac{1}{z}\frac{dz}{dx}=\frac{\sin x}{x}+\cos x\ln x$ .............. $\left(3\right)$

Substituting equation $\left(2\right)$ in equation $\left(3\right)$

$\frac{1}{x^{\sin x}}\frac{dz}{dx}=\frac{\sin x}{x}+\cos x\ln x$

$\frac{dz}{dx}=x^{\sin x}\frac{\sin x}{x}+x^{\sin x}\ln x$

$\frac{dz}{dx}=\sin x.x^{\sin x-1}+x^{\sin x}\ln x$ ......... $\left(4\right)$

Substituting equation $\left(4\right)$ in equation $\left(1\right)$

$\frac{dy}{dx}=\sin x^{\sin x-1}+\ln x,x^{\sin x}+\frac{1}{2\sqrt{x}}$