Question

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

Answer 1

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.