If f(x) is invertible with inverse g(y) then g ' (y) is

(a) 1/f(g(y))

(b)1/ f ' (g))

(c)1/f(x)

(d) undeined

(e) none of the above

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- October 16th 2006, 09:37 AMbobby77derivative of an invertible function
If f(x) is invertible with inverse g(y) then g ' (y) is

(a) 1/f(g(y))

(b)1/ f ' (g))

(c)1/f(x)

(d) undeined

(e) none of the above - October 16th 2006, 10:47 AMtopsquark
The inverse function theorem says that if y = f(x) has an inverse x = g(y), then then g'(y) will be 1/f'(x). There is an easy way to see this (though it is much more difficult to prove!):

If we have a function y(x), then dx/dy = 1/(dy/dx). You can "clear the fraction" in the "denominator" and see that the RHS is also dx/dy. By construction of the derivative, dx/dy would be the derivative of the inverse function.

Before someone more Mathematically minded than me comes along to remind you, let me say that treating dy/dx as a fraction is a*very*tricky process and is not a strictly legitimate process even where you can get away with it. (But it does occasionally provide a convenient guideline.)

-Dan - October 16th 2006, 05:38 PMThePerfectHacker
I never, ever, never use it. You can determine which class a person belongs to: a low class (physics) if he uses that trick, a high class (math) if he does not.

---

The complete proof is not so easy, it relys on properties of the interval.

Anyway here is an almost complete proof.

Let f be a differenciable function continous on the open interval (a,b).

Assume, f has an inverse differenciable function g.

If and only if,

f o g=x on (a,b)

Then the derivative is (chain rule: both are differenciable),

g' * f' o g=1

Note that f' o g is never zero on (a,b).

Equivalently,

g' = 1/(f' o g)

~~~

Why is this proof incomplete?

Actually this proof is complete, I did not prove that if f is differenciable and bijective on (a,b) then so too must be g, I assumed that if that is true then that is the case.

But that is not really such a bother if you are working with a "well-behave" invertible function then all you need to show that its inverse is differenciable and thus apply this proof. - October 17th 2006, 06:30 AMtopsquark
- October 17th 2006, 07:18 AMThePerfectHacker