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Math Help - Differential calculus problem

  1. #1
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    Differential calculus problem

    I am studying calculus at home as a hobby. I have 2 books: The Complete Idiot's Guide to Calculus by W. M. Kelley (K), and Calculus Made Easy (1998)by S.P.Thomson and M.Gardner (T).

    On page 119 of (K) a formula is derived which seems to show that the derivative of the inverse function equals the reciprocal of the derivative of the inverse function! This is contradicted by (T) page 148: "...the reciprocal of the derivative of the inverse function gives the derivative of the function itself. The latter version seems to me to be correct, but I cannot see where the other version goes wrong when it is produced in the "Critical Point " box on p119 of (K).

    It may be that I am missing the point because of the 2 different notations which are being used. Unfortunately I cannot access on my webpage any maths symbols, hence the description above. I would be grateful for an explanation of my problem and also some advice as to how I access maths symbols.
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  2. #2
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    Quote Originally Posted by Biggles View Post
    I am studying calculus at home as a hobby. I have 2 books: The Complete Idiot's Guide to Calculus by W. M. Kelley (K), and Calculus Made Easy (1998)by S.P.Thomson and M.Gardner (T).

    On page 119 of (K) a formula is derived which seems to show that the derivative of the inverse function equals the reciprocal of the derivative of the inverse function! This is contradicted by (T) page 148: "...the reciprocal of the derivative of the inverse function gives the derivative of the function itself. The latter version seems to me to be correct, but I cannot see where the other version goes wrong when it is produced in the "Critical Point " box on p119 of (K).

    It may be that I am missing the point because of the 2 different notations which are being used. Unfortunately I cannot access on my webpage any maths symbols, hence the description above. I would be grateful for an explanation of my problem and also some advice as to how I access maths symbols.
    f\left(f^{-1}(x)\right)=x\implies f'\left(f^{-1}(x)\right)\cdot \left(f^{-1}(x)\right)'=1...this can alternatively be written as \left(f^{-1}(x)\right)'=\frac{1}{f'\left(f^{-1}(x)\right)}
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  3. #3
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    Quote Originally Posted by Biggles View Post
    "the derivative of the inverse function equals the reciprocal of the derivative of the inverse function!"
    There is an excess "inverse" in that sentence. It should read: "The derivative of the inverse function equals the reciprocal of the derivative of the function." In symbols (as in Mathstud28's comment): \left(f^{-1}(x)\right)'=\frac{1}{f'\left(f^{-1}(x)\right)}.

    It may be easier to remember that in this alternative form: If y is an invertible differentiable function of x then \frac{dx}{dy} = \frac1{\frac{dy}{dx}}.
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  4. #4
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    Quote Originally Posted by Opalg View Post
    It may be easier to remember that in this alternative form: If y is an invertible differentiable function of x then \frac{dx}{dy} = \frac1{\frac{dy}{dx}}.
    Yes - the dash notation sometimes struggles to indicate the variable of differentiation. So, Biggles you might have assumed that

    (f^{-1}(x))'=f'\ (f^{-1}(x))

    (?) - if so then hence your question. Whereas,

    f'\ (f^{-1}(x)) is really meant to say "the derivative of \ \ f\ (f^{-1}(x)) with respect to f^{-1}(x)\ ."
    Last edited by tom@ballooncalculus; January 4th 2009 at 05:30 AM. Reason: typo
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  5. #5
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    Differential Calculus problem

    Dear Folks

    Thanks for answers, which have made the maths clear to me.

    However, I still have a small problem with the dash notation, but I cannot spell this out without acess to a maths symbol list. How do you write maths on this website - you all clearly have access to a symbol list?

    Biggles (maths tyro)
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  6. #6
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    Quote Originally Posted by Biggles View Post
    Dear Folks

    Thanks for answers, which have made the maths clear to me.

    However, I still have a small problem with the dash notation, but I cannot spell this out without acess to a maths symbol list. How do you write maths on this website - you all clearly have access to a symbol list?

    Biggles (maths tyro)
    http://www.mathhelpforum.com/math-help/latex-help/
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  7. #7
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    Quote Originally Posted by Biggles View Post
    I still have a small problem with the dash notation, but I cannot spell this out without acess to a maths symbol list. How do you write maths on this website - you all clearly have access to a symbol list?
    For that, you'll have to come to terms with LaTeX. Don't be put off if it looks intimidating at first it's not as hard as it seems.
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  8. #8
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    Differential Calculus problem

     Is (f^-1)' (x) the same as f'(f^-1(x)) ?
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  9. #9
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    Quote Originally Posted by Biggles View Post
    Is  (f^{-1})' (x) the same as f'(f^{-1}(x)) ?
    No. In fact,  (f^{-1})' (x) = \frac1{f'(f^{-1}(x))}. To see that, start from the fact that f(f^{-1}(x)) = x and differentiate both sides with respect to x, using the chain rule. That gives the equation f'(f^{-1}(x))(f^{-1})' (x) = 1.
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  10. #10
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    Differential Calculus problem

    Dear Folks

    After reading all the posted replies, and worrying for days over the problem of the apparent conflict between the 2 different versions of the equation, I confess that I am totally baffled!

    \frac{dy}{dx}=\frac{1}{\frac{dx}{dy}} seems very clear, and is equal to: " The derivative of the function is equal to the reciprocal of the derivative of the inverse function."

    If that is all correct, then I cannot understand the alternative version in the dash notation: (f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))}.
    This version seems to say: "The derivative of the inverse function is equal to the reciprocal of the derivative of the inverse function", which seems to contradict the first version, and indeed contradict itself!

    I can see how both versions are derived, but have been unable to sort out the apparent contradiction, which must I think be due to an error in my understanding of the dash notation. I think I must be misunderstanding the meaning of the left hand side of the dash equation.

    Please bear with my inability to resolve this problem.

    Biggles
    Last edited by mr fantastic; January 12th 2009 at 05:47 AM. Reason: Fixed the latex
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  11. #11
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    Quote Originally Posted by Biggles View Post
    I cannot understand the alternative version in the dash notation: (f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))}.
    This version seems to say: "The derivative of the inverse function is equal to the reciprocal of the derivative of the inverse function", which seems to contradict the first version, and indeed contradict itself!
    No, it says "The derivative of the inverse function (evaluated at x) is equal to the reciprocal of the derivative of the function [not the inverse function](evaluated at f^{-1}(x))."
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  12. #12
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    Hi Biggles! Opalg's way of putting it is perfect, but I hope you won't be hindered by this additional comment...

    I think you were thinking that

    f'(stuff )

    means the same as

    (f(stuff ))'

    However, the first one means the derivative of
    f(stuff ) with respect to stuff (some arithmetic or other stuff done to x, i.e. some function of x), and the second means the derivative of f(stuff ) with respect to x.

    The trouble, maybe, for a newcomer, is that the two do come to the same thing if stuff just is x.
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  13. #13
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    Dear Folks

    Many thanks to all for trying to make my problem witht the dash notation a little clearer. Special thanks to tom@balooncalculus for getting to the nub of my difficulty.

    The books that I have been studying have left something out in their efforts to make things "easy". I think I would be better off using a more formal approach. Please can someone recommend a book that makes the dash notation clear?

    Finally, congratulations to all for a great site.

    Biggles
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