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
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
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.
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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.