Question

The differential coefficient of $\sec \left({\tan }^{-1}\left(x\right)\right)$.

Answer 1

Determine the differential coefficient of the function.

The function $\sec \left({\tan }^{-1}\left(x\right)\right)$ is equivalent to the expression :

$$\begin{gathered} \sec \left({\tan }^{-1}\left(x\right)\right)=\sec \left({\sec }^{-1}\left(\sqrt{x^2+1}\right)\right)\left[\because {\tan }^{-1}\left(x\right)={\sec }^{-1}\left(\sqrt{x^2+1}\right)\right] \\ =\sqrt{x^2+1} \end{gathered}$$

The function is a compound function.

Determine the derivative:

$\begin{array}{rcl}\frac{d}{dx}\left(\sqrt{x^2+1}\right)& =& \frac{1}{2\sqrt{x^2+1}}\frac{d}{dx}\left(x^2+1\right)\\ & \Rightarrow & \frac{d}{dx}\left(\sqrt{x^2+1}\right)=\frac{1}{2\sqrt{x^2+1}}\left(2x\right)\\ & \Rightarrow & \frac{d}{dx}\left(\sqrt{x^2+1}\right)=\frac{x}{\sqrt{x^2+1}}\end{array}$

Hence, the differential coefficient is $\begin{array}{rcl}& & \frac{x}{\sqrt{x^2+1}}\end{array}$.