Chapter 6.3, Problem 10QR

To determine

To find: $\frac{dy}{dx}$

Answer to Problem 10QR

  $\frac{dy}{dx}=\frac{1}{1+x^2}$

Explanation of Solution

Given information: $y={\tan }^{-1}x$

Calculation:

  $y={\tan }^{-1}x$

Differentiating both sides of the equation with respect to x ,

  $\begin{array}{l}\frac{d}{dx}(y)=\frac{d}{dx}({\tan }^{-1}x)\\ \Rightarrow \frac{dy}{dx}=\frac{1}{1+x^2}\frac{dy}{dx}[\because \frac{d}{dx}({\tan }^{-1}u)=\frac{1}{1+u^2}\frac{du}{dx}]\\ \Rightarrow \frac{dy}{dx}=\frac{1}{1+x^2}\end{array}$

Thus, $\frac{dy}{dx}=\frac{1}{1+x^2}$