# What is the derivative of y=arctan(4x)?

Jul 31, 2014

$\frac{4}{16 {x}^{2} + 1}$

Explanation
First recall that $\frac{d}{\mathrm{dx}} \left[\arctan x\right] = \frac{1}{{x}^{2} + 1}$.

Via the chain rule:

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

2.) $\frac{d}{\mathrm{dx}} \left[\arctan 4 x\right] = \frac{4}{16 {x}^{2} + 1}$

If it isn't clear why $\frac{d}{\mathrm{dx}} \left[\arctan x\right] = \frac{1}{{x}^{2} + 1}$, continue reading, as I'll walk through proving the identity.

We will begin simply with

1.) $y = \arctan x$.

From this it is implied that

2.) $\tan y = x$.

Using implicit differentiation, taking care to use the chain rule on $\tan y$, we arrive at:

3.) ${\sec}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}} = 1$

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

4.) $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{{\sec}^{2} y}$

Which further simplifies to:

5.) $\frac{\mathrm{dy}}{\mathrm{dx}} = {\cos}^{2} y$

Next, a substitution using our initial equation will give us:

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

This might not look too helpful, but there is a trigonometric identity that can help us.

Recall ${\tan}^{2} \alpha + 1 = {\sec}^{2} \alpha$. This looks very similar to what we have in step 6. In fact, if we replace $\alpha$ with $\arctan x$, and rewrite the $\sec$ in terms of $\cos$ then we obtain something pretty useful:

${\tan}^{2} \left(\arctan x\right) + 1 = \frac{1}{{\cos}^{2} \left(\arctan x\right)}$

This simplifies to:

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

Now, simply multiply a few things around, and we get:

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

Beautiful. Now we can simply substitute into the equation we have in step 6:

7.) $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{{x}^{2} + 1}$

And voilà - there's our identity.