Question

What is the derivative of f (x) = tan^-1 (x) ?

Answer 1

Determine the derivative of $f\left(x\right)={\tan }^{-1}\left(x\right)$.

Consider $y={\tan }^{-1}\left(x\right)$.

$\Rightarrow \tan \left(y\right)=x\cdots \left(1\right)$

Differentiate the equation $\left(1\right)$ with respect to $x$.

$\begin{array}{rcl}\frac{d}{dx}\left(\tan \left(y\right)\right)& =& \frac{d}{dx}\left(x\right)\\ & \Rightarrow & \frac{d}{dy}\left(\tan \left(y\right)\right)·\frac{dy}{dx}=1\end{array}$

Apply the formula $\frac{d}{dx}\left(\tan \left(x\right)\right)={\sec }^2\left(x\right)$.

$\begin{array}{rcl}{\sec }^2\left(y\right)·\frac{dy}{dx}& =& 1\cdots \left(2\right)\end{array}$

Apply the trigonometric identity ${\sec }^2\left(y\right)=1+{\tan }^2\left(y\right)$.

Substitute the value to $\tan \left(y\right)$ into the trigonometric identity.

$\begin{array}{rcl}{\sec }^2\left(y\right)& =& 1+x^2\end{array}$

Substitute the value of ${\sec }^2\left(y\right)$ into $\left(2\right)$.

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

Hence, the derivative of the given function is $\begin{array}{rcl}\frac{dy}{dx}& =& \frac{1}{1+x^2}\end{array}$.