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## AP®︎/College Calculus AB

### Unit 3: Lesson 4

Differentiating inverse functions

# Derivatives of inverse functions

Functions f and g are inverses if f(g(x))=x=g(f(x)). For every pair of such functions, the derivatives f' and g' have a special relationship. Learn about this relationship and see how it applies to 𝑒ˣ and ln(x) (which are inverse functions!).

## Want to join the conversation?

• At , Sal says that y=e^x and x=e^y. This doesn't seem to make sense, I tried to plug some numbers into the calculator and the results don't match. If I am understanding this wrong, then please tell me where I went wrong, thanks. •  At , Sal says that, to get the inverse of y = eˣ, you swap the variables to get x = eʸ. If you graph the two functions y = eˣ and x = eʸ, you'll see they are symmetric around the function y = x, so they are indeed the inverse functions of each other.
• Will someone please explain how to go from y=e^x to x=e^y algebraically? • Is there a neat graphical or geometrical proof to this formula or is it only derived algebraically. I really want to see some beautiful math behind it • I'm not really sure how 1/e^x comes from g'(f(x)) at around , can anyone explain that for me, please? • I'm still stuck on the chain rule concept, shown again here. We've learned previously that the notation d/dx [g(x)] is the same as g'(x). So here, we see that d/dx [g(f(x)] is actually g'(f(x))*f'(x). So, d/dx [g(x)] should simultaneously be g'(f(x)) AND g'(f(x))*f'(x). I get that somehow the phrase 'with respect to...' plays a part here, but in this video, there's no indication that g'(f(x)) is 'with respect to' anything in particular. If someone could help me with this, I feel like it's all downhill from here!
(1 vote) • My left ear loved the video ;-) • Hi, I was trying to solve this question and was wondering if anyone knew how to approach it?

d/dx f^−1(4) where f(x) = 4 + 2x^3 + sin (πx/2)
for −1 ≤ x ≤ 1.
(1 vote) • So the goal is to evaluate d/dx(f^-1(x)) at x=4.

So f'(x) = 6x^2 + (pi/2)cos([pi/2]x))

Now the question is at what point should the derivative be evaluated. The key thing to note is the coordinates of x and y are swapped for the inverse.

So the x-coordinate for the inverse is 4 however the coordinate is swapped. So the for non-inverse function y=4.

So now the x-coordinate needs to be found for f(x)=4.

=> 4 = 4 + 2x^3 + sin(pi(x)/2)

=> 2x^3 + sin(pi(x)/2) = 0. By inspection x = 0 satisfies the equation.

Alternatively substitute x=4 for the inverse function then find the y-coordinate. The inverse function is

x = 4 + 2y^3 + sin((pi/2)y)

=> 0 = 2y^3 + sin((pi/2)y) since x=4.

Therefore y=0.

So the coordinate for the inverse function is (4, 0) and the non-inverse function (0, 4)

So you choose evaluate the expression using inverse or non-inverse function

Using f'(x) substituting x=0 yields pi/2 as the gradient.

=> d/dx f^-1(4) = (pi/2)^-1 = 2/pi since the coordinates of x and y are swapped.

So the gradient is given by

lim(dx->0) (f(x+dx) -f(x))/( (x+dx)-x )

However the x and y coordinates are swapped so the gradient for the inverse according differentiation by first principles is

lim(dx->0) ( (x+dx)-x ) / (f(x+dx) -f(x))

in others words the gradient of the inverse function is the reciprocal of the gradient of the non-inverse function
__________________
Alternative solution

From the inverse function: x = 4 + 2y^3 + sin((pi/2)y)

d/dx f^-1(x) =>
1 = 6y^2(dy/dx) + (pi/2)cos([pi/2]y)(dy/dx) (1)

This dy/dx next to each y(in equation (1)) comes from implicit differentiation. This is just a result from chain rule. If you want you can replace y with u and then apply chain rule and you will get the same result.

Equation (1) => 1 = ( 6y^2 + (pi/2)cos([pi/2]y) )dy/dx

=> 1/( 6y^2 + (pi/2)cos([pi/2]y) ) = dy/dx

substituting y=0 yields 2/pi for the derivative for the inverse function.
• What happens if `g'(f(x)) = 0`? Then we can't go from

`g'(f(x)) * f'(x) = 1`
to
`f'(x) = 1/g'(f(x))`

since we can't divide by 0 right? So what would f'(x) be in that case?
(1 vote) • So f^-1(x) should equal x, yes? if g(x) is then defined as the inverse of f(x), shouldn't g(x)=x and g(f(x))=f(x)? Can someone explain?
(1 vote) • I'm confused on the part in which he stated you could switch f'(x) and g'(x) towards the end of the video.
So, if I am given a point at f(4)=6, and the (line tangent is) f'(x) is -3/4x+9, considering that g(x) is f^-1(x), in other words the inverse, what is true about g'?
a. g'(4)=-4/3
b. g'(4)=-3/4
c. g'(6)=-4/3
d. g'(6)=-3/4
(1 vote) • The tangent and f'(x) are different. f'(x) gives the gradient for input x, while the tangent is a line that touches a point.

It is only reciprocal for the corresponding point.

Since (4, 6) is apart of f. That means (6, 4) is apart of g. This is because the coordinates of functions f and g are swapped.