Question

Find the derivative of $(\sin x{)}^{{\cos }^{-1}x}$ with respect to $x$.

Answer 1

$y=(\sin x{)}^{{\cos }^{-1}x}$

Taking log on both sides

$\log y={\cos }^{-1}x.\log \left(\sin x\right)$

Differentiating w.r.t x,

$\frac{1}{y}\left(\frac{dy}{dx}\right)={\cos }^{-1}x.\frac{d}{dx}\left[\log \right(\sin x\left)\right]+\log \left(\sin x\right)\frac{d}{dx}\left({\cos }^{-1}x\right)$

$={\cos }^{-1}x.[\frac{1}{\sin x}.\cos x]+log\left(\sin x\right).\left[\frac{-1}{\sqrt{1-x^2}}\right]$

$\frac{1}{y}\left(\frac{dy}{dx}\right)={\cos }^{-1}x.\cot x+\log \left(\sin x\right)\left[\frac{-1}{\sqrt{1-x^2}}\right]$

$[\because \frac{d}{dx}(logx)=\frac{1}{x}\frac{d}{dx}{\cos }^{-1}x=\frac{-1}{\sqrt{1-x^2}}]$

$\therefore \frac{dy}{dx}=y[{\cos }^{-1}x.\cot x+\log (\sin x).[\frac{-1}{\sqrt{1-x^2}}\left]\right]$

$\frac{dy}{dx}=\left(\sin x{)}^{{\cos }^{-1}x}\right[{\cos }^{-1}x.cotx-\frac{log\left(\sin x\right)}{\sqrt{1-x^2}}]$