Inverse Derivative

Question

For the function

𝑓(𝑥)=5+4𝑥^3

find the derivative of the inverse function 𝑓-1 at the point 𝑥=505.

(𝑓^-1)′(505)=   

Answer 1

The inverse function swaps x and y values from the original function.

So if (505, y) is a point on the inverse function, then (x, 505) is a point on the original function. To find that x value, let y = 505 in the original function definition.

505 = 5 + 4x³

4x³ = 500

x³=125

x=5

That means (5, 505) is the point on f and (505, 5) is the point on the inverse.

You could at this point actually find the definition of the inverse function, then find its derivative and evaluate at x = 505.

But there is a theorem stating that the slope of the inverse at x = 505 is equal to the reciprocal of the slope of the original at x = 5.

f'(x) = 12 x²

f'(5) = 12(25)=300

And the reciprocal is 1/300. That is the slope of the inverse function at x = 505.

Since it is possible to actually determine the formula for the inverse:

x = 5 + 4y³

4y³=x - 5

y³=(x-5)/4

y = [(x-5)/4]^1/3 (this is the definition of the inverse function, call it g(x)--note that g(505) = 5

g'(x) = 1/3 [(x-5)/4]^-2/3 (1/4) = 1/12 [4/(x-5)]^2/3

g'(505) = 1/12 [4/500]^2/3= 1/12 (1/25) = 1/300

Using the theorem on derivative on inverse function is a lot easier than actually finding the inverse function definition (which is sometimes not even possible).