Q. If $y = \log_{\sin x}{\cos{x,then\ \frac{dy}{dx}}}$ is equal to
A) $\frac{\tan{x\ log(\sin{x) - \cot{x\log{\cos x}}}}}{(\log{\sin{x\left. \ \ \right)^{2}}}}$
B) $\frac{{- tan}{x\ log\sin{x - \cot{x\log{\cos x}}}}}{(\log{\sin{x\left. \ \ \right)^{2}}}}$
C) $\frac{{- tan}{x\ log(\sin{x) + \cot{x\log{(\cos{x)}}}}}}{(\log{\sin{x\left. \ \ \right)^{2}}}}$
D) $\frac{\tan{x\ log(\sin{x) + \cot{\log{{(cos}{x)}}}}}}{\lbrack\log{{(sin}{x)\rbrack^{2}}}}$
Explanation
(b)
We have, $y = \log_{\sin x}{\cos x}$
$= \frac{\log{\cos x}}{\log{\sin x}}$
$= \frac{\log{\sin{x\frac{1}{\cos x}( - \sin{x) - \log{\cos{x\frac{1}{\sin x}\cos x}}}}}}{\left( \log{\sin x} \right)^{2}}$
$= \frac{- \tan{x\log{\sin{x - \cot{x\log{\cos x}}}}}}{\left( \log{\sin x} \right)^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.