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

Aug 26, 2015

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

#### Explanation:

You can differentiate a function $y = {\tan}^{- 1} \left({x}^{2}\right)$ by using implicit differentiation.

So, if you have a function $y = {\tan}^{- 1} \left({x}^{2}\right)$, then you know that you can write

$\tan \left(y\right) = {x}^{2}$

Differentiate both sides with respect to $x$ to get

$\frac{d}{\mathrm{dy}} \left(\tan y\right) \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left({x}^{2}\right)$

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

This is equivalent to saying that

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

Remember that you have

$\textcolor{b l u e}{{\sec}^{2} x = 1 + {\tan}^{2} x}$

which means that you get

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

Finally, replace ${\tan}^{2} y$ with ${x}^{2}$ to get

$\frac{\mathrm{dy}}{\mathrm{dx}} = \textcolor{g r e e n}{\frac{2 x}{1 + {x}^{4}}}$