Random math problem

Here’s a random problem from what we’re doing these days in calc. The solution is in the permalink.

f(x) = (tanx + arctanx)^4
Find f'(x)

f'(x) = 3(tanx + arctanx)^3 [sec^2(x) + 1/(1 +x^2)]

That’s all the work. It’s all definintions… Everything in the [‘s has had the derivative taken (just keeps things clear). For interested parties, here’s how it breaks down:

Since it’s raised to the fourth power we need to use the chain rule. We multiply the base by the power and then raise it to the power – 1. So at that point we get:

f'(x) = 3(tanx + arctanx)^3

The chain rule also states that we need to multiply that by the derivative of the base. The tricky part here is that we’re taking the derivative of arctan(x). It’s defined as: 1/(1 +x^2) which you see in the final answer.