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Thread: Derivative of inverse function

  1. #1
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    Derivative of inverse function

    I'm reading the proof for the following theorem.

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

    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

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

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

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

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

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

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

    Thanks a lot if you can help
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by acc100jt View Post
    I'm reading the proof for the following theorem.

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

    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

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

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

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

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

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

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

    Thanks a lot if you can help
    It wouldn't make any sense if $\displaystyle f^{-1}$ weren't differentiable, because then $\displaystyle g(x)$ would not be differentiable, and the entire discussion would be irrelevant. Do you see what I mean?
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  3. #3
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    not quite .

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

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

    Thanks
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  4. #4
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    the first proof in the textbook uses Newton's difference quotient to prove, but the condition that $\displaystyle f^{-1}$ is differentiable is not required, but the discussion is still relevant, why? i'm getting confused haha
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  5. #5
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by acc100jt View Post
    not quite .

    Referring to the theorem, the RESULT tells us that $\displaystyle f^{-1}$ is differentiable, whereas the second proof (the one in Leibniz notation) requires $\displaystyle f^{-1}$ to be differentiable, right?
    The "result" shows the relationship of the inverse functions as they relate in "calculus". If $\displaystyle f$ 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 $\displaystyle g'(a)=\frac{1}{f'[g(a)]}$

    Quote Originally Posted by acc100jt View Post
    Since $\displaystyle y=f^{-1}(x)$, so do you mean that when we are writing $\displaystyle \frac{dy}{dx}$, we are assuming that $\displaystyle y=f^{-1}(x)$ is differentiable?

    Thanks
    Yes, but it's not so much an assumption as a necessary condition...
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  6. #6
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    GREAT!!! all doubts cleared. Thanks a lot!!! VonNemo19
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