**Answer** ##4/(16x^2 + 1)##

**Explanation** First recall that ##d/dx[arctan x] = 1/(x^2 + 1)##.

Via the :

1.) ##d/dx[arctan 4x] = 4/((4x)^2 + 1)##

2.) ##d/dx[arctan 4x] = 4/(16x^2 + 1)##

If it isn’t clear why ##d/dx[arctan x] = 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 dy/dx = 1##

Solving for ##dy/dx## gives us:

4.) ##dy/dx = 1/(sec^2 y)##

Which further simplifies to:

5.) ##dy/dx = cos^2 y##

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

6.) ##dy/dx = cos^2(arctan x)##

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

Recall ##tan^2alpha + 1 = sec^2alpha##. 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(arctan x) + 1 = 1/(cos^2(arctan x))##

This simplifies to:

##x^2 + 1 = 1/(cos^2(arctan x))##

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

##1/(x^2 + 1) = cos^2(arctan x)##

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

7.) ##dy/dx = 1/(x^2 + 1)##

And voilĂ – there’s our identity.