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!).
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- At3:10, 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.(4 votes)
- At3:03, 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.(31 votes)
- Will someone please explain how to go from y=e^x to x=e^y algebraically?(2 votes)
- You can't, those two equations aren't equivalent.
If you meant go from y=e^x to x=ln(y), you just take the natural log of both sides and simplify ln(e^x) as x.(23 votes)
- 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(5 votes)
- I'm not really sure how 1/e^x comes from g'(f(x)) at around4:00, can anyone explain that for me, please?(2 votes)
- When plugging the value 'x' into g'(x), we get 1/x. Now, instead of 'x', we are plugging f(x) into the function (i.e 1/f(x)). You can then replace 'f(x)' with the actual function (in this case was 1/e^x), resulting in 1/(1/e^x).(5 votes)
- 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)
- Yes, one of the issues with Lagrange notation ("prime notation") is that it doesn't state what variable we are differentiating with respect to.
However, it is commonly accepted that
𝑔'(𝑥) = 𝑑∕𝑑𝑥 [𝑔(𝑥)]
𝑔'(𝑓(𝑥)) = 𝑑∕𝑑𝑓 [𝑔(𝑓(𝑥))]
𝑔(𝑓(𝑥))' = 𝑑∕𝑑𝑥 [𝑔(𝑓(𝑥))](4 votes)
- My left ear loved the video ;-)(2 votes)
- 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.
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
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.(3 votes)
- What happens if
g'(f(x)) = 0? Then we can't go from
g'(f(x)) * f'(x) = 1
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)
- If 𝑔'(𝑓(𝑥)) = 0, then 𝑔'(𝑓(𝑥))⋅𝑓 '(𝑥) = 0, which means that 𝑔(𝑓(𝑥)) is a constant, and 𝑓 and 𝑔 are not each other's inverses.(2 votes)
- 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 hope I understood your question.
Pay attention to the domains, g(x) and f(x) can not both be in X, one must be the inverse.
f = g^-1 and f^-1 = g s.t. g(f(x)) = x
In your example f^-1(x) and g(x) are identical, that does not work.
g(f(x)) = x not g^-1(f(x))
g(f(x)) = f(x), then g is the identity function
not the inverse.
Hope this helps(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'?
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.
Assuming the gradient at x=4 is -3/4, the gradient is -4/3.
If you meant f'(x) = -3/4x + 9, then g'(6) = 16/141.(1 vote)
- [Instructor] So let's say I have two functions that are the inverse of each other. So I have f of x, and then I also have g of x, which is equal to the inverse of f of x. And f of x would be the inverse of g of x as well. If the notion of an inverse function is completely unfamiliar to you, I encourage you to review inverse functions on Khan Academy. Now, one of the properties of inverse functions are that if I were to take g of f of x, g of f of x, or I could say the f inverse of f of x, that this is just going to be equal to x. And it comes straight out of what an inverse of a function is. If this is x right over here, the function f would map to some value f of x. So that's f of x right over there. And then the function g, or f inverse, if you input f of x into it, it would take you back, it would take you back to x. So that would be f inverse, or we're saying g is the same thing as f inverse. So all of that so far is a review of inverse functions, but now we're going to apply a little bit of calculus to it, using the chain rule. And we're gonna get a pretty interesting result. What I want to do is take the derivative of both sides of this equation right over here. So let's apply the derivative operator, d/dx on the left-hand side, d/dx on the right-hand side. And what are we going to get? Well, on the left-hand side, we would apply the chain rule. So this is going to be the derivative of g with respect to f of x. So that's going to be g prime of f of x, g prime of f of x, times the derivative of f of x with respect to x, so times f prime of x. And then that is going to be equal to what? Well, the derivative with respect to x of x, that's just equal to one. And this is where we get our interesting result. All we did so far is we used something we knew about inverse functions, and we'd use the chain rule to take the derivative of the left-hand side. But if you divide both sides by g prime of f of x, what are you going to get? You're going to get a relationship between the derivative of a function and the derivative of its inverse. So you get f prime of x is going to be equal to one over all of this business, one over g prime of f of x, g prime of f of x. And this is really neat because if you know something about the derivative of a function, you can then start to figure out things about the derivative of its inverse. And we can actually see this is true with some classic functions. So let's say that f of x is equal to e to the x, and so g of x would be equal to the inverse of f. So f inverse, which is, what's the inverse of e to the x? Well, one way to think about it is, if you have y is equal to e to the x, if you want the inverse, you can swap the variables and then solve for y again. So you'd get x is equal to e to the y. You take the natural log of both sides, you get natural log of x is equal to y. So the inverse of e to the x is natural log of x. And once again, that's all review of inverse functions. All right, if that's unfamiliar, review it on Khan Academy. So g of x is going to be equal to the natural log of x. Now, let's see if this holds true for these two functions. Well, what is f prime of x going to be? Well, this one of those amazing results in calculus. One of these neat things about the number e is that the derivative of e to the x is e to the x. And in other videos, we also saw that the derivative of the natural log of x is one over x. So let's see if this holds out. So we should get a result, f prime of x, e to the x should be equal to one over g prime of f of x. So g prime of f of x, so g prime is one over our f of x, and f of x is e to the x, one over e to the x. Is this indeed true? Yes, it is. One over, one over e to the x is just going to be e to the x. So it all checks out. And you could do the other way because these are inverses of each other. You could say g prime of x is going to be equal to one over f prime of g of x because they're inverses of each other. And actually, what's really neat about this, is that you could actually use this to get a sense of what the derivative of an inverse function is even going to be.