Continuity and Differentiability — Class 12 Maths Solution

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

therefore,$\frac{d}{{dy}}\log y \cdot \frac{{dy}}{{dx}} = \frac{d}{{dx}}(\cos x \cdot \log \sin x)$

$\Rightarrow$ $\frac{1}{y} \cdot \frac{{dy}}{{dx}} = \cos x \cdot \frac{d}{{dx}}\log \sin x + \log \sin x \cdot \frac{d}{{dx}}\cos x$

$= \cos x \cdot \frac{1}{{\sin x}} \cdot \frac{d}{{dx}}\sin x + \log \sin x \cdot ( - \sin x)$

$= \cot x \cdot \cos x - \log (\sin x) \cdot \sin x$

therefore,$\frac{{dy}}{{dx}} = y\left[ {\frac{{{{\cos }^2}x}}{{\sin x}} - \sin x \cdot \log (\sin x)} \right]$

$= \sin {x^{\cos x}}\left[ {\frac{{{{\cos }^2}x}}{{\sin x}} - \sin x \cdot \log (\sin x)} \right]$

$\Rightarrow \frac{dy}{dx} = \sin {x^{\cos x}}\left[ {\frac{{{{\cos }^2}x}}{{\sin x}} - \sin x \cdot \log (\sin x)} \right]$