Let $y=log_e\left(\frac{a+bsinx}{a-bsinx}\right),$ then value of $\frac{dy}{dx}$ is :
Answer & explanation
Correct answer: option 4
The correct answer is option (4) → $\frac{2ab\cos x}{a^2-b^2\sin^2x}$
$y=\log_e\left(\frac{a+b\sin x}{a-b\sin x}\right)$
$\frac{dy}{dx}=\frac{a-b\sin x}{a+b\sin x}\frac{(b\cos x(a-b\sin x)+b\cos x(a+b\sin x))}{(a-b\sin x)^2}$
$=\frac{2ab\cos x}{a^2-b^2\sin^2x}$