# How do you take the derivative of tan^-1 2x?

Aug 4, 2015

Use the derivative of ${\tan}^{-} 1$ and the chain rule.

#### Explanation:

The derivative of ${\tan}^{-} 1 x$ is $\frac{1}{1 + {x}^{2}}$ (for "why", see note below)

So, applying the chain rule, we get:

$\frac{d}{\mathrm{dx}} \left({\tan}^{-} 1 u\right) = \frac{1}{1 + {u}^{2}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

In this question $u = 2 x$, so we get:

$\frac{d}{\mathrm{dx}} \left({\tan}^{-} 1 2 x\right) = \frac{1}{1 + {\left(2 x\right)}^{2}} \cdot \frac{d}{\mathrm{dx}} \left(2 x\right)$

$= \frac{2}{1 + 4 {x}^{2}}$

Note

If $y = {\tan}^{-} 1 x$, then $\tan y = x$

Differentiating implicitly gets us:

${\sec}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}} = 1 , \text{ }$ so

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\sec} ^ 2 y$

From trigonometry, we know that $1 + {\tan}^{2} y = {\sec}^{2} y$

so $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{1 + {\tan}^{2} y}$

and we have $\tan y = x$, so we get:

For $y = {\tan}^{-} 1 x$, the derivative is:

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