Question

Find $f^{\prime }\left(x\right)$ if $f\left(x\right)=(\sin x{)}^{\sin x}$ for all $0<x<\pi$.

Answer 1

Let $y=(\sin x{)}^{\sin x}$

Taking $\text{log}$ on both sides, we get
$\text{log}y=\text{log}\left(\sin x^{\sin x}\right)$

$\text{log}y=\sin x.\text{log}\left(\sin x\right)$

$(\because \text{log}(a^n)=n.\text{log}a)$

$\frac{d\left(\text{log}y\right)}{dx}=\frac{d(\sin x.\text{log}(\sin x\left)\right)}{dx}$

$\frac{d\left(\text{log}y\right)}{dy}\left(\frac{dy}{dx}\right)=\frac{d(\sin x.\text{log}(\sin x\left)\right)}{dx}$

Differentiating both sides w.r.t. $x$ , we get,

$\frac{1}{y}.\frac{dy}{dx}=\frac{d(\sin x.\text{log}(\sin x\left)\right)}{dx}$

Using Product Rule : $(uv{)}^{\prime }=u^{\prime }v+v^{\prime }u$

$\frac{1}{y}.\frac{dy}{dx}=\frac{d(\sin x.)}{dx}.\text{log}\left(\sin x\right)+\frac{d\left(\text{log}\right(\sin x\left)\right)}{dx}.\sin x$

$\frac{1}{y}.\frac{dy}{dx}=\cos x\text{log}\left(\sin x\right)+\frac{1}{\sin x}.\cos x.\sin x$

$\frac{1}{y}.\frac{dy}{dx}=\cos x\text{log}\left(\sin x\right)+\cos x$

$\frac{dy}{dx}=y(\cos x.\text{log}(\sin x)+\cos x)$

Substituting $y=(\sin x{)}^{\sin x}$

$\frac{dy}{dx}=\left(\sin x{)}^{\sin x}\right(\cos x.\text{log}\left(\sin x\right)+\cos x)$

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

$\Rightarrow f^{\prime }\left(x\right)=(\sin x{)}^{\sin x}.\cos x(\text{log}\left(\sin x\right)+1)$