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

1 Answer
Aug 21, 2014

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.