Question

Find derivative of $f\left(x\right)=\frac{\cos x}{1+\sin x}$
[Hint: ${\sin }^2\theta +{\cos }^2\theta =1$]

• A. $\frac{1}{(1+\sin x)}$
• B. $\frac{-1}{(1+\sin x)}$
• C. $\frac{-1}{(1-\sin x)}$
• D. $\frac{-1}{(1-\cos x)}$

Answer 1

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

Given $f\left(x\right)=\frac{\cos x}{1+\sin x}$
By using Division rule we have derivative as
$f^{{}^{\prime }}\left(x\right)=\frac{(1+\sin x)(\cos x{)}^{{}^{\prime }}-(1+\sin x{)}^{{}^{\prime }}\left(\cos x\right)}{(1+\sin x{)}^2}$
$=\frac{(1+\sin x)(-\sin x)-\cos x\left(\cos x\right)}{(1+\sin x{)}^2}$
$=\frac{-\sin x-{\sin }^{2}x-{\cos }^{2}x}{(1+\sin x{)}^2}$
$=\frac{-\sin x-({\sin }^{2}x+{\cos }^{2}x)}{(1+\sin x{)}^2}$
$=\frac{-\sin x-1}{(1+\sin x{)}^2}=\frac{-(\sin x+1)}{(1+\sin x{)}^2}$
$=\frac{-1}{(1+\sin x)}$