Question

If $f\left(x\right)=\frac{\cos x}{1+\sin x}$, then $f^{\prime }\left(x\right)$

• A. $\frac{1}{1-\sin x}$
• B. $\frac{-1}{1+\sin x}$
• C. $\frac{1+\sin x}{x}$
• D. None

Answer 1

The correct option is B $\frac{-1}{1+\sin x}$

Given, $f\left(x\right)=\frac{\cos x}{1+\sin x}$
$\therefore f^{\prime }\left(x\right)=\frac{-\sin x(1+\sin x)-\cos x.\cos x}{(1+\sin x{)}^2}$
$=-\frac{[\sin x+{\sin }^{2}x+{\cos }^{2}x]}{(1+\sin x{)}^2}$
$=\frac{-(\sin x+1)}{(1+\sin x{)}^2}$

$=\frac{-1}{1+\sin x}$