# What is the derivative of  (arctan x)^3?

Nov 14, 2017

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

#### Explanation:

We are given $y = {\arctan}^{3} \left(x\right)$ and need to find $\frac{\mathrm{dy}}{\mathrm{dx}}$.

By the power rule, if $\textcolor{red}{y = {x}^{a}}$, then, $\textcolor{red}{\frac{\mathrm{dy}}{\mathrm{dx}} = a {x}^{a - 1}}$.

Unfortunately, we can't directly apply the power rule to find $\frac{\mathrm{dy}}{\mathrm{dx}}$ here as the base is $\arctan \left(x\right)$, not $x$.

Instead, the power rule can only find the derivative with respect to $\arctan \left(x\right)$, or $\frac{\mathrm{dy}}{d \left(\arctan \left(x\right)\right)} = 3 {\arctan}^{2} \left(x\right)$.

However, by the chain rule, we can multiply both sides by $\textcolor{red}{\frac{d \left(\arctan \left(x\right)\right)}{\mathrm{dx}}}$ to get $\frac{\mathrm{dy}}{\mathrm{dx}} = 3 {\arctan}^{2} \left(x\right) \frac{d \left(\arctan \left(x\right)\right)}{\mathrm{dx}}$.

(Note: the left hand side cancels out: $\frac{\mathrm{dy}}{\cancel{d \left(\arctan \left(x\right)\right)}} \cdot \frac{\cancel{d \left(\arctan \left(x\right)\right)}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dx}}$.)

Now, we just need to find $\frac{d \left(\arctan \left(x\right)\right)}{\mathrm{dx}}$.

Consider the function $u = \arctan \left(x\right)$. This necessarily means that $\tan \left(u\right) = x$.

Now, if we differentiate both sides, we will get $\frac{1}{\cos} ^ 2 \left(u\right) = \frac{\mathrm{dx}}{\mathrm{du}}$, or $\frac{\mathrm{du}}{\mathrm{dx}} = {\cos}^{2} \left(u\right)$.

We said previously that $u = \arctan \left(x\right)$. Substitute this in to find that $\frac{\mathrm{du}}{\mathrm{dx}} = {\cos}^{2} \left(\arctan \left(x\right)\right) = \frac{1}{{\tan}^{2} \left(\arctan \left(x\right)\right) + 1} = \frac{1}{{x}^{2} + 1}$.

Note: ${\cos}^{2} \left(\arctan \left(x\right)\right)$ is simplified using the identity ${\tan}^{2} \left(\theta\right) + 1 = \frac{1}{\cos} ^ 2 \left(\theta\right)$ (found by dividing both sides of ${\sin}^{2} \left(\theta\right) + {\cos}^{2} \left(\theta\right) = 1$ by ${\cos}^{2} \left(\theta\right)$ and using the fact that $\tan \left(\theta\right) = \sin \frac{\theta}{\cos} \left(\theta\right)$).
This identity can be arranged to ${\cos}^{2} \left(\theta\right) = \frac{1}{{\tan}^{2} \left(\theta\right) + 1}$.

Now, we have $\frac{d \left(\arctan \left(x\right)\right)}{\mathrm{dx}} = \frac{1}{{x}^{2} + 1}$.

Since $\frac{\mathrm{dy}}{\mathrm{dx}} = 3 {\arctan}^{2} \left(x\right) \frac{d \left(\arctan \left(x\right)\right)}{\mathrm{dx}}$, our final answer is $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 {\arctan}^{2} \left(x\right)}{{x}^{2} + 1}$.