# How do you differentiate arctan(x^2)?

##### 2 Answers
Jun 6, 2016

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

#### Explanation:

The derivative of the arctangent function is:

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

So, when applying the chain rule, this becomes

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

So, for $\arctan \left({x}^{2}\right)$, where $f \left(x\right) = {x}^{2}$, we have

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

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

Jun 6, 2016

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

#### Explanation:

Let $y = \arctan \left({x}^{2}\right)$.

Then $\tan \left(y\right) = {x}^{2}$. From here, differentiate both sides of the equation. Recall that the chain rule comes into effect on the left-hand side.

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

Dividing both sides by ${\sec}^{2} \left(y\right)$, which is equivalent to multiplying both sides by ${\cos}^{2} \left(y\right)$, gives

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\cos}^{2} \left(\arctan \left({x}^{2}\right)\right) \cdot 2 x$

Note that ${\cos}^{2} \left(\arctan \left({x}^{2}\right)\right)$ can be simplified.

If $\arctan \left({x}^{2}\right)$ is an angle in a right triangle, then ${x}^{2}$ is the side opposite the angle and $1$ is the side adjacent to the angle. Then, by the Pythagorean Theorem, $\sqrt{1 + {x}^{4}}$ is the triangle's hypotenuse.

Since cosine is the adjacent side, $1$, divided by the hypotenuse, $\sqrt{1 + {x}^{4}}$, we see that $\cos \left(\arctan \left({x}^{2}\right)\right) = \frac{1}{\sqrt{1 + {x}^{4}}}$.

Therefore

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