The derivative of $\tan ^{-1}\left[\frac{x}{1+\sqrt{1-x^2}}\right]$ with respect to $\sec ^{-1}\left(\frac{1}{2 x^2-1}\right)$ is

Question

Aicharya · Accepted Answer

Let $\begin{aligned} y & =\tan ^{-1}\left(\frac{x}{1+\sqrt{1-x^2}}\right) \ u & =\sec ^{-1}\left(\frac{1}{2 x^2-1}\right) \end{aligned}$ Let $x=\cos \theta$ \(\begin{aligned} & y=\tan ^{-1}\left(\frac{\cos \theta}{1+\sin \theta}\right), u=\sec ^{-1}(\sec 2 \theta) \ & \Rightarrow \quad y=\tan ^{-1}\left(\frac{1}{\sec \theta+\tan \theta}\right),(u=2 \theta) \ & \Rightarrow \quad y=\tan ^{-1}(\sec \theta-\tan \theta),(u=2 \theta) \ & \Rightarrow \tan y=\sec \theta-\tan \theta, u=2 \theta \ & \Rightarrow \sec ^2 y \cdot \frac{d y}{d \theta}=\sec \theta \cdot \tan \theta-\sec ^2 \theta, \frac{d u}{d \theta}=2 \ & \Rightarrow \quad \frac{d y}{d \theta}=\frac{\sec \theta \cdot \tan \theta-\sec ^2 \theta}{\sec ^2 y}, \frac{d u}{d \theta}=2 \ & \Rightarrow \quad \frac{d y}{d \theta}=\frac{\sec \theta \cdot \tan \theta-\sec ^2 \theta}{1+(\sec \theta-\tan \theta)^2}, \frac{d u}{d \theta}=2 \ & \Rightarrow \quad \frac{d y}{d u}=\frac{1}{2}\left[\frac{\sec \theta \cdot \tan \theta-\sec ^2 \theta}{1+\sec ^2 \theta+\tan ^2 \theta-2 \sec \theta \cdot \tan \theta}\right] \ & =\frac{1}{2} \frac{\sec \theta(\tan \theta-\sec \theta)}{2 \sec ^2 \theta-2 \sec \theta \cdot \tan \theta} \Rightarrow \frac{d y}{d u}=-\frac{1}{4} \ \end{aligned}\)

The derivative of \(\tan ^{-1}\left[\frac{x}{1+\sqrt{1-x^2}}\right]\) with respect to \(\sec…

Solution