Question

${\displaystyle \frac{\text{d}}{\text{dx}}\left[{\left(\cos x\right)}^{\log x}\right]}$ = ______.

Options

• ${\displaystyle {\left(\cos x\right)}^{\log x}(\frac{1}{x}\log \cos x+\tan x)}$

• ${\displaystyle {\left(\cos x\right)}^{\log x}(\frac{1}{x}\log \cos x-\tan x\log x)}$

• ${\displaystyle {\left(\cos x\right)}^{\log x}(\frac{1}{x}\log \cos x+\log x)}$

• ${\displaystyle {\left(\cos x\right)}^{\log x}(x\log \cos x-\tan x\log x)}$

Answer 1

${\displaystyle \frac{\text{d}}{\text{dx}}\left[{\left(\cos x\right)}^{\log x}\right]=\underline{{\left(\cos x\right)}^{\log x}(\frac{1}{x}\log \cos x-\tan x\log x)}}$.

Explanation:

Let y = ${\displaystyle {\left(\cos x\right)}^{\log x}}$

Taking logarithm on both sides, we get

log y = log x log (cos x)

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

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

${\displaystyle =\frac{\text{dy}}{\text{dx}}={\left(\cos x\right)}^{\log x}(\frac{1}{x}\log \cos x-\tan x\log x)}$