Question

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

• A. $\frac{[\cot x\mathrm{logcos}x+\tan x\mathrm{logsin}x)]}{(\mathrm{logcos}x{)}^2}$
• B. $\frac{[\tan x\mathrm{logcos}x+\cot x\mathrm{logsin}x)]}{(\mathrm{logcos}x{)}^2}$
• C. $\frac{[\cot x\mathrm{logcos}x+\tan x\mathrm{logsin}x)]}{(\mathrm{logsin}x{)}^2}$
• D. none of these

Answer 1

The correct option is A

$\frac{[\cot x\mathrm{logcos}x+\tan x\mathrm{logsin}x)]}{(\mathrm{logcos}x{)}^2}$

Explanation for the correct option.

Find the $\frac{dy}{dx}$

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

Using the property of logarithmic, ${\log }_{b}a=\frac{\log a}{\log b}$.

$y=\frac{\mathrm{logsin}x}{\mathrm{logcos}x}$

Now using the quotient rule for differentiation i.e; $\frac{d\left({\displaystyle \frac{u}{v}}\right)}{dx}=\frac{v{\displaystyle \frac{du}{dx}}-u{\displaystyle \frac{dv}{dx}}}{v^2}$
$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{[\mathrm{logcos}x\times \left(\frac{1}{\sin x}\right)\times \cos x-\mathrm{logsin}x\times \left({\displaystyle \frac{1}{\cos x}}\right)\times (-\sin x\left)\right]}{(\mathrm{logcos}x{)}^2}\\ & =& \frac{[\cot x\mathrm{logcos}x+\tan x\mathrm{logsin}x)]}{(\mathrm{logcos}x{)}^2}\end{array}$
Hence option A is correct.