# How do you find the derivative of the function: (arctan(6x^2 +5))^2?

Jan 23, 2016

$\frac{24 x \arctan \left(6 {x}^{2} + 5\right)}{{\left(6 {x}^{2} + 5\right)}^{2} + 1}$

#### Explanation:

This will require a little chain rule.

The first issue is the second power, which can be dealt with as such: $\frac{d}{\mathrm{dx}} \left[{u}^{2}\right] = 2 u \cdot u '$. We know that $u = \arctan \left(6 {x}^{2} + 5\right)$, so

$f ' \left(x\right) = 2 \arctan \left(6 {x}^{2} + 5\right) \frac{d}{\mathrm{dx}} \left[\arctan \left(6 {x}^{2} + 5\right)\right]$

Now, we must deal with the $\arctan$ function, which uses the rule $\frac{d}{\mathrm{dx}} \left[\arctan \left(u\right)\right] = \frac{u '}{{u}^{2} + 1}$, and we have $u = 6 {x}^{2} + 5$, giving us

$f ' \left(x\right) = 2 \arctan \left(6 {x}^{2} + 5\right) \cdot \frac{\frac{d}{\mathrm{dx}} \left[6 {x}^{2} + 5\right]}{{\left(6 {x}^{2} + 5\right)}^{2} + 1}$

Simplify:

$f ' \left(x\right) = \frac{2 \arctan \left(6 {x}^{2} + 5\right) \cdot 12 x}{{\left(6 {x}^{2} + 5\right)}^{2} + 1}$

$f ' \left(x\right) = \frac{24 x \arctan \left(6 {x}^{2} + 5\right)}{{\left(6 {x}^{2} + 5\right)}^{2} + 1}$