Question

What is the derivative of `(arctan x)^3`?

Answer 1

`dy/dx=(3arctan^2(x))/(x^2+1)`

Explanation:

We are given `y=arctan^3(x)` and need to find `dy/dx`.

By the power rule, if `color(red)(y=x^a)`, then, `color(red)(dy/dx=ax^(a-1))`.

Unfortunately, we can't directly apply the power rule to find `dy/dx` here as the base is `arctan(x)`, not `x`.

Instead, the power rule can only find the derivative with respect to `arctan(x)`, or `dy/(d(arctan(x)))=3arctan^2(x)`.

However, by the chain rule, we can multiply both sides by `color(red)((d(arctan(x)))/dx)` to get `dy/dx=3arctan^2(x)(d(arctan(x)))/dx`.

(Note: the left hand side cancels out: `dy/cancel(d(arctan(x)))*cancel(d(arctan(x)))/dx=dy/dx`.)

Now, we just need to find `(d(arctan(x)))/dx`.

Consider the function `u=arctan(x)`. This necessarily means that `tan(u)=x`.

Now, if we differentiate both sides, we will get `1/cos^2(u)=dx/(du)`, or `(du)/dx=cos^2(u)`.

We said previously that `u=arctan(x)`. Substitute this in to find that `(du)/dx=cos^2(arctan(x))=1/(tan^2(arctan(x))+1)=1/(x^2+1)`.

Note: `cos^2(arctan(x))` is simplified using the identity `tan^2(theta)+1=1/cos^2(theta)` (found by dividing both sides of `sin^2(theta)+cos^2(theta)=1` by `cos^2(theta)` and using the fact that `tan(theta)=sin(theta)/cos(theta)`).
This identity can be arranged to `cos^2(theta)=1/(tan^2(theta)+1)`.

Now, we have `(d(arctan(x)))/dx=1/(x^2+1)`.

Since `dy/dx=3arctan^2(x)(d(arctan(x)))/dx`, our final answer is `dy/dx=(3arctan^2(x))/(x^2+1)`.