Derivative of inverse function

**acc100jt** · November 7th 2009, 07:38 PM

I'm reading the proof for the following theorem.

If f is a one-to-one differentiable function with inverse function $g=f^{-1}$ and $f'(g(a)) \neq 0$ , then the inverse function is differentiable at a and
$g'(a)=\frac{1}{f'(g(a))}$

The Calculus textbook that I'm reading gives two proofs, the first one uses Newton's difference quotient, which I can understand.

The 2nd goes like this:

Replacing a by the general number x in the theorem, we get

$g'(x)=\frac{1}{f'(g(x))}$

If we write $y=g(x)$ , then $f(y)=x$ , so the equation can be expressed in Leibniz notation.

$\frac{dy}{dx}=\frac{1}{\frac{dx}{dy}}$

IF IT IS KNOWN IN ADVANCE THAT $f^{-1}$ IS DIFFERENTIABLE, then its derivative can be computed more easily by using implicit differentiation.

$f(y)=x$
$f'(y)\frac{dy}{dx}=1$
$\frac{dy}{dx}=\frac{1}{f'(y)}=\frac{1}{\frac{dx}{d y}}$

My question is this, I don't understand which step of the proof requires the fact that $f^{-1}$ is differentiable

Thanks a lot if you can help

**VonNemo19** · November 7th 2009, 07:43 PM

Originally Posted by acc100jt

I'm reading the proof for the following theorem.

If f is a one-to-one differentiable function with inverse function $g=f^{-1}$ and $f'(g(a)) \neq 0$ , then the inverse function is differentiable at a and
$g'(a)=\frac{1}{f'(g(a))}$

The Calculus textbook that I'm reading gives two proofs, the first one uses Newton's difference quotient, which I can understand.

The 2nd goes like this:

Replacing a by the general number x in the theorem, we get

$g'(x)=\frac{1}{f'(g(x))}$

If we write $y=g(x)$ , then $f(y)=x$ , so the equation can be expressed in Leibniz notation.

$\frac{dy}{dx}=\frac{1}{\frac{dx}{dy}}$

IF IT IS KNOWN IN ADVANCE THAT $f^{-1}$ IS DIFFERENTIABLE, then its derivative can be computed more easily by using implicit differentiation.

$f(y)=x$
$f'(y)\frac{dy}{dx}=1$
$\frac{dy}{dx}=\frac{1}{f'(y)}=\frac{1}{\frac{dx}{d y}}$

My question is this, I don't understand which step of the proof requires the fact that $f^{-1}$ is differentiable

Thanks a lot if you can help

It wouldn't make any sense if $f^{-1}$ weren't differentiable, because then $g(x)$ would not be differentiable, and the entire discussion would be irrelevant. Do you see what I mean?

**acc100jt** · November 7th 2009, 08:15 PM

not quite

.

Referring to the theorem, the RESULT tells us that $f^{-1}$ is differentiable, whereas the second proof (the one in Leibniz notation) requires $f^{-1}$ to be differentiable, right?

Since $y=f^{-1}(x)$ , so do you mean that when we are writing $\frac{dy}{dx}$ , we are assuming that $y=f^{-1}(x)$ is differentiable?

Thanks

**acc100jt** · November 7th 2009, 08:19 PM

the first proof in the textbook uses Newton's difference quotient to prove, but the condition that $f^{-1}$ is differentiable is not required, but the discussion is still relevant, why? i'm getting confused haha

**VonNemo19** · November 7th 2009, 08:22 PM

Originally Posted by acc100jt

not quite

.

Referring to the theorem, the RESULT tells us that $f^{-1}$ is differentiable, whereas the second proof (the one in Leibniz notation) requires $f^{-1}$ to be differentiable, right?

The "result" shows the relationship of the inverse functions as they relate in "calculus". If $f$ is one to one and differentiable (smooth), it necessarily follows that its inverse is differentiable and smooth. It is a statement meant to be taken at face value. Note that this, however, is not what the proof is trying to show. The proof is trying to show that $g'(a)=\frac{1}{f'[g(a)]}$

Originally Posted by acc100jt

Since $y=f^{-1}(x)$ , so do you mean that when we are writing $\frac{dy}{dx}$ , we are assuming that $y=f^{-1}(x)$ is differentiable?

Thanks

Yes, but it's not so much an assumption as a necessary condition...

**acc100jt** · November 7th 2009, 08:33 PM

GREAT!!! all doubts cleared. Thanks a lot!!! VonNemo19

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