How do you Find the derivative of #y=arctan(x^2)#?

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

Here is slightly different - we have an #x^2# instead of an #x#.

Via the chain rule:

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

2.) #d/dx[arctan x^2] = (2x)/(x^4 + 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 voilà - there's our identity.