Question

Sin x power cos x differentiation

Answer 1

$\frac{dy}{dx} = {sinx}^{cosx}\left ( \frac{cos^{2}{}x}{sinx} - sinx\times\log{x} \right )$

Step-by-step explanation:

given,

$y = {sinx}^{cosx}$

Taking log on both side,

$log\times(y) = log\times({sinx})^{cosx}$

$log\times(y) = cos{x}\times\ln{sinx}$

Using 'chain rule' to differentiate,$\ln{sinx}$

$\frac{1}{y}\times{dy} = cos{x}\times\frac{1}{sinx}{\times{cosx}} + log{sinx}{\times(-sinx)}dx$

$\frac{dy}{dx} = y\times\left ( cos{x}\times\frac{1}{sinx}{\times{cosx}} + log{sinx}{\times(-sinx)} \right )$

$\frac{dy}{dx} = {sinx}^{cosx}\left ( \frac{cos^{2}{}x}{sinx} - sinx\times\log{x} \right )$