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Derivative of inverse function.

  • Thread starter Willowz
  • Start date
  • #1
184
1

Homework Statement


Let, f(x) = x^3 - 3x^2 - 1 , x => 2 . Find the value of df^-1/dx at the point x=-1=f(3)


Homework Equations


The definition states; "If f has an interval I as domain and f'(x) exists and is never zero on I, then f'(c) is diff at every point in its domain. The value of (f^-1)' at a point b in the domain of f_1 is the reciprocal of the value of f1 at the point a = f^-1(b):
b80fffc5e854e0c44c3e6accbfadf7a2.png

Though I don't really understand what is going on in the equation.

The Attempt at a Solution


The derivative is f' = 3x^2 - 6x , but I don't really get what and why I do what I have to do from here. Coulden't find a youtube video for these sort of problems.
 

Answers and Replies

  • #2
Char. Limit
Gold Member
1,204
13
Wow, that definition sounds complicated. The definition I always used is, well... y = f(x), yeah? So the inverse of f(x) would be x(y), and the derivative of that is dx/dy. Now it might seem intuitive, and there are pitfalls in that, but this is true because of the chain rule:

[tex]\left(f^{-1}(x)\right)' = \frac{dx}{dy} = \frac{1}{\frac{dy}{dx}}[/tex]

So the derivative of the inverse of your function is the reciprocal of the derivative of your function. Fun stuff, really.
 
  • #3
184
1
Wow, that definition sounds complicated. The definition I always used is, well... y = f(x), yeah? So the inverse of f(x) would be x(y), and the derivative of that is dx/dy. Now it might seem intuitive, and there are pitfalls in that, but this is true because of the chain rule:

[tex]\left(f^{-1}(x)\right)' = \frac{dx}{dy} = \frac{1}{\frac{dy}{dx}}[/tex]

So the derivative of the inverse of your function is the reciprocal of the derivative of your function. Fun stuff, really.
Oh, ok. So, could you tell me what does the question ask for when they ask to find df^-1/dx at x=-1=f(3) ?
 
  • #4
Deveno
Science Advisor
906
6
it makes more sense if you write it this way:

dx/dy = 1/(dy/dx), but this isn't very accurate notation.

in english: the derivative of the inverse, is the reciprocal of the derivative.

in other words, instead of writing y = f(x), you consider x = f-1(y).

your equation should read:

[tex](f^{-1})'(f(a)) = \frac{1}{f'(a)}[/tex] or

[tex](f^{-1})'(b) = \frac{1}{f'(f^{-1}(b))}[/tex]

since presumably b = f(a) (and thus a = f-1(b)).
 
  • #5
184
1
it makes more sense if you write it this way:

dx/dy = 1/(dy/dx), but this isn't very accurate notation.

in english: the derivative of the inverse, is the reciprocal of the derivative.

in other words, instead of writing y = f(x), you consider x = f-1(y).

your equation should read:

[tex](f^{-1})'(f(a)) = \frac{1}{f'(a)}[/tex] or

[tex](f^{-1})'(b) = \frac{1}{f'(f^{-1}(b))}[/tex]

since presumably b = f(a) (and thus a = f-1(b)).
So, in this case I don't have to find the inverse of the function because x=-1=f(3);
[tex](f^{-1})'(b) = \frac{1}{f'(f^{-1}(3))}[/tex] ----> [tex](f^{-1})'(b) = \frac{1}{f'(-1)}[/tex] ?
 

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