Question

If $e^y=\frac{e^{\sin x}(1+\cos x{)}^n}{(1-\cos x{)}^m}$, find $\frac{dy}{dx}$.

• A. $\frac{dy}{dx}=\cos x-n\frac{\sin x}{1+\cos x}-m\frac{\sin x}{1-\cos x}$
• B. $\frac{dy}{dx}=\cos x-n\frac{\sin x}{1-\cos x}-m\frac{\sin x}{1-\cos x}$
• C. $\frac{dy}{dx}=-\cos x-n\frac{\sin x}{1+\cos x}-m\frac{\sin x}{1-\cos x}$
• D. $\frac{dy}{dx}=-\cos x-n\frac{\sin x}{1-\cos x}-m\frac{-\sin x}{1-\cos x}$

Answer 1

The correct option is A $\frac{dy}{dx}=\cos x-n\frac{\sin x}{1+\cos x}-m\frac{\sin x}{1-\cos x}$

$e^y=e^{\sin x}\frac{(1+\cos x{)}^n}{(1-\cos x{)}^m}$.

Taking log of both sides,

$y=\sin x+n\log (1+\cos x)-m\log (1-\cos x)$

$\therefore \frac{dy}{dx}=\cos x-n\frac{\sin x}{1+\cos x}-m\frac{\sin x}{1-\cos x}$