Derivative Inverse Question

Question

For each of the given functions f(x) , find the derivative (f^-1)'(c) at the given point c , first finding a = f^-1(c)

1. f(x) = 3x+7x^15 ; c = -10

a = -1

(f^-1)'(c) = ????

1. f(x) = x^2 - 15x + 77 on the interval [7.5, infinity); c=23

a = ????

(f^-1)'(c) = ????

Answer 1

This seems to be checking your understanding of the following

if you have a function f(x) and it has an inverse f^-1(x), then

(f^-1)'(a) = 1/f'(c)

where f(a) = c and f^-1(c) = a

I will go through the 1st question assuming that, and hopefully you will be able to answer the 2nd question on your own after that.

1) f(x) = 3x+7x^15 ; c = -10 a= -1 we want to find (f^-1)'(10)

We are told that f^-1(c) = a so we can use (f^-1)'(a) = 1/f'(c) then

(f^-1)'(-10) = 1/f'(-1)

We need to first evaluate f'(-1) to evaluate (f^-1)'(-10)

we need to derive f'(x)

f(x) = 3 + 7x¹⁵ so using the power rule for derivatives we get:

f'(x) = 3 + 7(15)x¹⁴ = 3 + 105x¹⁴

evaluating at x=-1

f'(-1) = 3 + 105(-1)¹⁴ -1 raised to an even power is 1

f(-1) = 3 + 105 = 108

going back and plugging into (f^-1)'(-10) = 1/f'(-1)

We get the answer

(f^-1)'(-10) = 1/108

Optional in case you are confused about (f^-1)'(a) = 1/f'(c)

Below I go through one of the ways you can derive this by using the chain rule

Recall for that if you have an inverse of a function f(x), f^-1(x) then

f(a) = c means f^-1(c) = a

or in general f^-1(f(x)) = x

so if we take the derivative with respect to x on both sides we have:

d/dx[f^-1(f(x))] = d/dx[x]

using the chain rule on the left side we get

d/dx [f^-1(f(x)] f'(x) = 1

so

d/dx f^-1(f(x)) = 1/f'(x)

In particular if x= a, then we have

d/dx f^-1(f(a)) = 1/f'(a)

but if f(a) = c then

(f^-1)'(c) = 1/f'(a)