Question

If $y={\log }_{\cos x}\sin x$, then $\frac{dy}{dx}$ is equal to

• A. $\frac{(\cot x\log \cos x+\tan x\log \sin x)}{{\left(\log \cos x\right)}^2}$
• B. $\frac{(\tan x\log \cos x+\cot x\log \sin x)}{{\left(\log \cos x\right)}^2}$
• C. $\frac{(\cot x\log \cos x+\tan x\log \sin x)}{{\left(\log \sin x\right)}^2}$
• D. None of the above

Answer 1

Answer: (A)

$\frac{(\cot x\log \cos x+\tan x\log \sin x)}{{\left(\log \cos x\right)}^2}$

Given, $y={\log }_{\cos x}\sin x=\frac{\log \sin x}{\log \cos x}$

On differentiating w.r.t. $x$, we get

$\frac{dy}{dx}=\frac{\cot x.\log \cos x+\tan x.\log \sin x}{{\left(\log \cos x\right)}^2}$