Question

How do you differentiate `arctan(x^2)`?

Answer 1

`(2x)/(1+x^4)`

Explanation:

The derivative of the arctangent function is:

`d/dxarctan(x)=1/(1+x^2)`

So, when applying the chain rule, this becomes

`d/dxarctan(f(x))=1/(1+(f(x))^2)*f'(x)`

So, for `arctan(x^2)`, where `f(x)=x^2`, we have

`d/dxarctan(x^2)=1/(1+(x^2)^2)*d/dx(x^2)`

`=(2x)/(1+x^4)`

Answer 2

`(2x)/sqrt(1+x^4)`

Explanation:

Let `y=arctan(x^2)`.

Then `tan(y)=x^2`. From here, differentiate both sides of the equation. Recall that the chain rule comes into effect on the left-hand side.

`sec^2(y)*dy/dx=2x`

Dividing both sides by `sec^2(y)`, which is equivalent to multiplying both sides by `cos^2(y)`, gives

`dy/dx=cos^2(y)*2x`

`dy/dx=cos^2(arctan(x^2))*2x`

Note that `cos^2(arctan(x^2))` can be simplified.

If `arctan(x^2)` is an angle in a right triangle, then `x^2` is the side opposite the angle and `1` is the side adjacent to the angle. Then, by the Pythagorean Theorem, `sqrt(1+x^4)` is the triangle's hypotenuse.

Since cosine is the adjacent side, `1`, divided by the hypotenuse, `sqrt(1+x^4)`, we see that `cos(arctan(x^2))=1/sqrt(1+x^4)`.

Therefore

`dy/dx=(1/sqrt(1+x^4))^2*2x=1/(1+x^4)*2x=(2x)/(1+x^4)`