If $u= an ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x} ight)$ and $v= an ^{-1}\left(\frac{2 x \sqrt{1-x^{2}}}{1-2 x^{2}} ight)$, then $\frac{d u}{d v}$ at $x=0$ is

Question

If $u=	an ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}ight)$ and $v=	an ^{-1}\left(\frac{2 x \sqrt{1-x^{2}}}{1-2 x^{2}}ight)$, then $\frac{d u}{d v}$ at $x=0$ is

Aicharya · Accepted Answer

Given $u=	an ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}ight)$ Put $x=	an 	heta$ $u=	an ^{-1}\left(\frac{\sec 	heta-1}{	an 	heta}ight) \quad=	an ^{-1}\left(\frac{1-\cos 	heta}{\sin 	heta}ight)=	an ^{-1}\left(\frac{2 \sin ^{2} \frac{	heta}{2}}{2 \sin \frac{	heta}{2} \cos }ight)$ $	herefore u=\frac{	an ^{-1} x}{2} \Rightarrow \frac{d u}{d x}=\frac{1}{2\left(1+x^{2}ight)}$ We have, $v=	an ^{-1}\left(\frac{2 x \sqrt{1-x^{2}}}{1-2 x^{2}}ight)$ Put $x=\sin 	heta$ $	herefore v=	an ^{-1}\left(\frac{2 \sin 	heta \cdot \cos 	heta}{\cos 2 	heta}ight)=	an ^{-1}\left(\frac{\sin 2 	heta}{\cos 2 	heta}ight)=	an ^{-1}(	an 2 	heta)=2 	heta$ $	herefore v=2 	an ^{-1} x \Rightarrow \frac{d v}{d x}=\frac{2}{1+x^{2}}$ $\left.\frac{d u}{d v}=\frac{d u}{d x}ight)=\frac{1}{2\left(1+x^{2}ight)} 	imes \frac{\left(1+x^{2}ight)}{2}=\frac{1}{4}$

If $u=\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)$ and $v=\tan ^{-1}\left(\frac{2 x…

Solution