Continuity and Differentiability — Class 12 Maths Solution

Let $y = {(\log x)^{\cos x}}$ ...(i)

By taking log on both sides of (i), we get

$\log y = \cos x\log (\log x)$ ….(ii)

On differentiating (ii) both sides w.r.t. x, we get

$\cfrac{1}{y}\cfrac{{dy}}{{dx}} = \cos x\cfrac{d}{{dx}}\log (\log x) + \log (\log x)\cfrac{d}{{dx}}\cos x$

$= \cos x \cdot \cfrac{1}{{\log x}}\cfrac{1}{x} + \log (\log x)( - \sin x)$

$= \cfrac{{\cos x}}{{x\log x}} - \sin x\log (\log x)$

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