Question

How do you differentiate `f(x) = 1/sqrt(arctan(e^(x-1))` using the chain rule?

Answer 1

`f'(x) = frac{-e^(x-1)}{2((e^(x-1))^2+1)sqrt((arctan(e^(x-1)))^3)}`

Explanation:

Let `f(x) = [arctan(e^(x-1))]^(-1/2)`

Use the chain rule, which says that `color(blue)(d/dx(f(g(x))) = f'(g(x))*g'(x))`:

`f'(x) = (-1/2)(arctan(e^(x-1)))^(-3/2)*(frac{1}{(e^(x-1))^2+1})*e^(x-1)`

Now we simplify to get:
`f'(x) = frac{-e^(x-1)}{2((e^(x-1))^2+1)sqrt((arctan(e^(x-1)))^3)}`