Question

Question about the derivative of an inverse function; details in the reply.

Answer 1

If f is continuous and monotonous over the closed interval [a,b] and f'(x) is different from zero for every x in [a,b], then the derivative of the inverse function x = f^-1(y) is given by
\[ \frac{dx}{dy} = \frac{1}{dy/dx} \]
I must show that the formula above can be written as:
\[ (f^{-1})'(x) = -\frac{ 1 }{ f'(f^{-1}(x)) } \]
I begin by noticing that dx/dy is the derivative of the inverse function, so it can be written as:
\[ \frac{dx}{dy}=(f^{-1})'(x) \]
Now, for the desired formula to appear, it appears that dx/dy (which is the derivative of the function f) should be:
\[ f'(x) = f'(f^{-1}(x)) \]
but I'm having trouble making sense out of it.
It appears that x is being replace by f^-1(x), and I suppose it is so because x is the image of the inverse function, and so is f^-1(x). But it still seems strange, because it should be x = f^-1(y), and not f^-1(x). Writing x = f^-1(x) doesn't seem to make any sense.

Answer 2

The thing is if \( y = f(x) \), then in terms of the same variables, \( x = f^{-1}(y) \).

Hence you should write
\[ \frac{dx}{dy} = (f^{-1})'(y)\]
Not as a function if x.

Answer 3

So I would have the expression below?
\[f'(x) = f'(f^{-1}(y))\]

Answer 4

Yes.

Answer 5

Thus in terms of the same variables:
\[(f^{-1})'(y) = -\frac{ 1 }{ f'(f^{-1}(y)) }\]

Answer 6

Yes

Answer 7

Then I can argue that writing the expression above would be equivalent to express it using x instead of y?
\[(f^{-1})'(x) = -\frac{ 1 }{ f'(f^{-1}(x)) }\]

Answer 8

Sure, because x and y are just dummy variables.

Answer 9

OK, very good. I think I get it now.

Answer 10

Just one more question, would that be correct to write the expression below?
\[ (f^{-1})'(y) = \frac{1}{f'(x)} \]