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

1 Answer
Jun 10, 2018

Answer:

#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)}#