Let #f# and #g# be functions that are differentiable everywhere. If #g# is the inverse function of #f# and if #g(-2) = 5# and #f'(5) = -1/2#, then what is #g'(-2)#?

1 Answer
Jan 15, 2018

#g'(-2) = -2#

Explanation:

If #g(x)# is the inverse function of #f(x)#, it follows that:

#f(g(x)) = x#

Differentiate the identity using the chain rule:

#d/dx f(g(x)) = d/dx x#

#f'(g(x)) g'(x) = 1#

So:

#g'(x) = 1/(f'(g(x))#

For #x= -2# we have:

#g'(-2) = 1/(f'(g(-2))) = 1/(f'(5)) = 1/(-1/2) = -2#