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    So  \displaystyle \frac{d}{dx} f^{-1}(x) = \frac{1}{f'(f^{-1}(x))}

    I understand this can be shown from the identity map \displaystyle f \circ f^{-1}(x) = x and just differentiating and making  \displaystyle \frac{d}{dx} f^{-1}(x) the subject.

    But this looked familiar when my lecture notes said it's the same as  \displaystyle \frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} if you let  \displaystyle y = f^{-1}(x)

    Now I'm probably being stupid but, wouldn't that make the above  \displaystyle \frac{dy}{dx} = \frac{1}{\frac{d}{dx}y}= \frac{1}{\frac{dy}{dx}}

 

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    Can you explain further as I see it, if:

     \displaystyle y = f^{-1}(x) \Leftrightarrow f(y) = x.

    Differentiate with respect to y:

     \displaystyle \frac{dx}{dy} = f^{'} (y) = f^{'} \Big( f^{-1} (x) \Big).

    Therefore:

     \displaystyle \frac{dy}{dx} = \frac{1}{f^{'} \Big( f^{-1} (x) \Big)}, \, \Bigg( = \frac{ d[f^{-1} (x)] }{dx} \Bigg).

    ------------------------------------------------------------------------------------------------

    Off topic: Is this to do with the "Inverse Function Theorem"?
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    (Original post by simon0)
    Can you explain further as I see it, if:

     \displaystyle y = f^{-1}(x) \Leftrightarrow f(y) = x.

    Differentiate with respect to y:

     \displaystyle \frac{dx}{dy} = f^{'} (y) = f^{'} \Big( f^{-1} (x) \Big).

    Therefore:

     \displaystyle \frac{dy}{dx} = \frac{ d[f^{-1} (x)] }{dx} = \frac{1}{f^{'} \Big( f^{-1} (x) \Big)}.

    ------------------------------------------------------------------------------------------------

    Off topic: Is this to do with the "Inverse Function Theorem"?
    How did you get the last line From what's above?

    It might be, but I'm just reading the rules for differentiation.
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    I believe you are missing the last part of the equation, the F and exponent can't just disappear and should be included, so you have 2/3's of the equation. Apologies if I am incorrect and on the wrong track, its been a while since I had quadratic equations, Calc 1 & 2, etc.

    (Original post by NotNotBatman)
    So  \displaystyle \frac{d}{dx} f^{-1}(x) = \frac{1}{f'(f^{-1}(x))}

    I understand this can be shown from the identity map \displaystyle f \circ f^{-1}(x) = x and just differentiating and making  \displaystyle \frac{d}{dx} f^{-1}(x) the subject.

    But this looked familiar when my lecture notes said it's the same as  \displaystyle \frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} if you let  \displaystyle y = f^{-1}(x)

    Now I'm probably being stupid but, wouldn't that make the above  \displaystyle \frac{dy}{dx} = \frac{1}{\frac{d}{dx}y}= \frac{1}{\frac{dy}{dx}}

 

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    (Original post by NotNotBatman)
    How did you get the last line From what's above?

    It might be, but I'm just reading the rules for differentiation.
    I applied the reciprocal of dx/dy (so I did 1/(dx/dy) ). :-)
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    (Original post by luq_ali)
    I believe you are missing the last part of the equation, the F and exponent can't just disappear and should be included, so you have 2/3's of the equation. Apologies if I am incorrect and on the wrong track, its been a while since I had quadratic equations, Calc 1 & 2, etc.
    The equation is written properly. I'm not sure what you mean... the f^-1 is there.
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    (Original post by simon0)
    I applied the reciprocal of dx/dy (so I did 1/(dx/dy) ). :-)
    But that's what need to be proven.
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    Yes, it is correct in the last one you had there, but I was referring to the last line of your original...I was writing my response and by the time I had completed, you had made a correction to your initial post-which is the one (the first one) I was responding too. So now it is indeed correct.

    (Original post by NotNotBatman)
    The equation is written properly. I'm not sure what you mean... the f^-1 is there.
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    (Original post by luq_ali)
    Yes, it is correct in the last one you had there, but I was referring to the last line of your original...I was writing my response and by the time I had completed, you had made a correction to your initial post-which is the one (the first one) I was responding too. So now it is indeed correct.
    I haven't edited my post, maybe it took while to load or something I don't know.

    But I still don't understand how my working is wrong :/

    it isn't correct as it would mean that dy/dx at some point is always equal to it's reciprocal.
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    The equation as stated by Simon is correct, and that is what I was noting-you quoted it in responding to him and asking about it-what he gave was the correct one as far as I can tell.

    (Original post by NotNotBatman)
    I haven't edited my post, maybe it took while to load or something I don't know.

    But I still don't understand how my working is wrong :/

    it isn't correct as it would mean that dy/dx at some point is always equal to it's reciprocal.
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    (Original post by luq_ali)
    The equation as stated by Simon is correct, and that is what I was noting-you quoted it in responding to him and asking about it-what he gave was the correct one as far as I can tell.
    Oh ok, but what I want to know is how what I've written is incorrect (concerning the relationship between notation), because it seems like the answer was used to prove the original question.
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    One short way to show that dy/dx = 1/(dy/dx) is using the chain rule where:

    y = f(x).

    Differentiate with respect to y using the chain rule:

    1 = d[f(x)] / dy = ( d[f(x)]/dx ) ( dx/dy) = (dy/dx)(dx/dy), as we defined y = f(x).

    Therefore:

    dy/dx = 1/( dx/dy ).

    This is assuming dx/dy is not 0.

    A more formal proof is using the inverse function theorem.
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    Your final equation in your last line-compare it what Simon has as his final line-and try to work back from there.





    (Original post by NotNotBatman)
    Oh ok, but what I want to know is how what I've written is incorrect (concerning the relationship between notation), because it seems like the answer was used to prove the original question.
 
 
 
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