# How do you find the derivative of y = arctan(x^2)?

Sep 16, 2016

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

#### Explanation:

Note that $\frac{d}{\mathrm{dx}} \arctan \left(x\right) = \frac{1}{1 + {x}^{2}}$. Thus, according to the chain rule:

$\frac{d}{\mathrm{dx}} \arctan \left(f \left(x\right)\right) = \frac{1}{1 + f {\left(x\right)}^{2}} \cdot f ' \left(x\right)$

Thus:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \arctan \left({x}^{2}\right) = \frac{1}{1 + {\left({x}^{2}\right)}^{2}} \cdot \frac{d}{\mathrm{dx}} {x}^{2} = \frac{2 x}{1 + {x}^{4}}$

Sep 16, 2016

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

#### Explanation:

Rearrange the equation:

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

Differentiate both sides. Recall to use the chain rule on the left hand side.

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

Note that ${\sec}^{2} \left(y\right) = {\tan}^{2} \left(y\right) + 1$:

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

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

$\left({x}^{4} + 1\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 2 x$

Solving for $\frac{\mathrm{dy}}{\mathrm{dx}}$:

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