Question

Find derivative of ${\tan }^{-1}\frac{\cos x}{1+\sin x}$.

• A. $\frac{1}{2}$
• B. $-\frac{1}{2}$
• C. $\frac{3}{2}$
• D. $-\frac{3}{2}$

Answer 1

The correct option is B $-\frac{1}{2}$

$\frac{d}{dx}\left(\arctan \left(\frac{\cos \left(x\right)}{1+\sin \left(x\right)}\right)\right)$

$=\frac{d}{du}\left(\arctan \left(u\right)\right)\frac{d}{dx}\left(\frac{\cos \left(x\right)}{1+\sin \left(x\right)}\right)$

We know that $\frac{d}{du}\left(\arctan \left(u\right)\right)=\frac{1}{u^2+1}$ and $\frac{d}{dx}\left(\frac{\cos \left(x\right)}{1+\sin \left(x\right)}\right)=\frac{\frac{d}{dx}\left(\cos \left(x\right)\right)(1+\sin \left(x\right))-\frac{d}{dx}(1+\sin \left(x\right))\cos \left(x\right)}{{(1+\sin \left(x\right))}^2}=\frac{(-\sin \left(x\right))(1+\sin \left(x\right))-\cos \left(x\right)\cos \left(x\right)}{{(1+\sin \left(x\right))}^2}$

$=\frac{1}{-1-\sin \left(x\right)}$

So, $\frac{d}{dx}\left(\arctan \left(\frac{\cos \left(x\right)}{1+\sin \left(x\right)}\right)\right)=\frac{1}{u^2+1}\cdot \frac{1}{-1-\sin \left(x\right)}=\frac{1}{{\left(\frac{\cos \left(x\right)}{1+\sin \left(x\right)}\right)}^2+1}\cdot \frac{1}{-1-\sin \left(x\right)}$

$=-\frac{\sin \left(x\right)+1}{{\cos }^2\left(x\right)+{(\sin \left(x\right)+1)}^2}$

Therefore, $f^{\prime }\left(0\right)=-\frac{1}{2}$