# What's the derivative of arctan(x^3/3)?

Feb 16, 2016

$\frac{9 {x}^{2}}{{x}^{6} + 9}$

#### Explanation:

We can use the chain rule, which states that

d/dx(f(g(x))=f'(g(x))*g'(x)

In the case of an arctangent function, it will help to know that

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

When we apply this to the chain rule, we see that

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

When differentiating $\arctan \left({x}^{3} / 3\right)$, we see that $g \left(x\right) = {x}^{3} / 3$, yielding a derivative of:

$\frac{d}{\mathrm{dx}} \left(\arctan \left({x}^{3} / 3\right)\right) = \frac{1}{{\left({x}^{3} / 3\right)}^{2} + 1} \cdot \frac{d}{\mathrm{dx}} \left({x}^{3} / 3\right)$

Through the power rule, we know that $\frac{d}{\mathrm{dx}} \left({x}^{3} / 3\right) = {x}^{2}$. The rest becomes simplification:

$= \frac{1}{{x}^{6} / 9 + 1} \cdot {x}^{2}$

$= {x}^{2} / \left(\frac{{x}^{6} + 9}{9}\right)$

$= \frac{9 {x}^{2}}{{x}^{6} + 9}$