Question

Find the derivative of $\frac{\cos x}{1+\sin x}.$

Answer 1

Let $f\left(x\right)=\frac{\cos x}{1+\sin x}$
Differentiating with respect to $x$
$\Rightarrow f^{\prime }\left(x\right)=\frac{d}{dx}\left(\frac{\cos x}{1+\sin x}\right)$
$\Rightarrow f^{\prime }\left(x\right)=\frac{(1+\sin x)\frac{d}{dx}\left(\cos x\right)-\left(\cos x\right)\frac{d}{dx}(1+\sin x)}{(1+\sin x{)}^2}$$\Rightarrow f^{\prime }\left(x\right)=\frac{(1+\sin x)(-\sin x)-\cos x\left(\cos x\right)}{(1+\sin x{)}^2}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{-\sin x(1+\sin x)-{\cos }^{2}x}{(1+\sin x{)}^2}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{-\sin x-{\sin }^{2}x-{\cos }^{2}x}{(1+\sin x{)}^2}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{-\sin x-({\sin }^{2}x+{\cos }^{2}x)}{(1+\sin x{)}^2}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{-\sin x-1}{(1+\sin x{)}^2}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{-(1+\sin x)}{(1+\sin x{)}^2}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{-1}{1+\sin x}$