The derivative of $\sec ^{-1}\left(\frac{1}{2 x^2-1}\right)$ with respect to $\sqrt{1-x^2}$ at $x=\frac{1}{2}$ equals

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Aicharya · Accepted Answer

Let $u=\sec ^{-1}\left(\frac{1}{2 x^2-1}ight)$ and $v=\sqrt{1-x^2}$ Put $x=\cos 	heta$ $	herefore \quad u=\sec ^{-1}\left(\frac{1}{2 \cos ^2 	heta-1}ight)$ and $v=\sqrt{1-\cos ^2 	heta}$ $\begin{aligned} & \Rightarrow u=\sec ^{-1}(\sec 2 	heta) 	ext { and } v=\sin 	heta \ & \Rightarrow u=2 	heta 	ext { and } v=\sin 	heta \ & \Rightarrow \frac{d u}{d 	heta}=2 	ext { and } \frac{d v}{d 	heta}=\cos 	heta \Rightarrow \frac{d u}{d v}=\frac{d u / d 	heta}{d v / d 	heta}=\frac{2}{\cos 	heta}\end{aligned}$ $\Rightarrow \frac{d u}{d v}=\frac{2}{x}$ $[\because \cos 	heta=x]$ At $\quad x=\frac{1}{2} \Rightarrow\left(\frac{d u}{d v}ight)_{x=1 / 2}=\frac{2}{1 / 2}=4$

The derivative of $\sec ^{-1}\left(\frac{1}{2 x^2-1}\right)$ with respect to $\sqrt{1-x^2}$ at…

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