Hey there! Sign in to join this conversationNew here? Join for free
    • Community Assistant
    • Thread Starter
    Offline

    18
    ReputationRep:
    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}}

 

?
    Offline

    10
    ReputationRep:
    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"?
    • Community Assistant
    • Thread Starter
    Offline

    18
    ReputationRep:
    (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.
    Offline

    21
    ReputationRep:
    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}}

 

?
    Offline

    10
    ReputationRep:
    (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) ). :-)
    • Community Assistant
    • Thread Starter
    Offline

    18
    ReputationRep:
    (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.
    • Community Assistant
    • Thread Starter
    Offline

    18
    ReputationRep:
    (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.
    Offline

    21
    ReputationRep:
    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.
    • Community Assistant
    • Thread Starter
    Offline

    18
    ReputationRep:
    (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.
    Offline

    21
    ReputationRep:
    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.
    • Community Assistant
    • Thread Starter
    Offline

    18
    ReputationRep:
    (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.
    Offline

    10
    ReputationRep:
    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.
    Offline

    21
    ReputationRep:
    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.
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    What newspaper do you read/prefer?
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.