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Math Help - Differentation question

  1. #1
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    Differentation question

    Hi guys. Struggling with some revision questions.
    So if you could step out how you solved the questions, that'd be amazing.

    firstly;

    y(m) = (sqrt m) ln m + e^m tan m
    (brackets just to show that the square root symbols only covers the m)

    and

    y + e^ t = m - e^(2m^2)

    please help!
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  2. #2
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    Re: Differentation question

    Just in case a picture helps...



    ... where (key in spoiler) ...

    Spoiler:


    ... is the product rule. Straight continuous lines differentiate downwards (integrate up) with respect to m. And,



    Spoiler:




    Second... is t a function of m? Full question please.


    _________________________________________

    Don't integrate - balloontegrate!

    Balloon Calculus; standard integrals, derivatives and methods

    Balloon Calculus Drawing with LaTeX and Asymptote!
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  3. #3
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    Re: Differentation question

    sorry typed the question in wrong. the question is y + e^y = x-e^2x^2

    so from your answer i should be getting dy/dm = 1/2 m^-(1/2) ln m + sqrt(m) 1/m + e^m tan m + e^m sec^2 m ?
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  4. #4
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    Re: Differentation question

    so from your answer i should be getting dy/dm = 1/2 m^-(1/2) ln m + sqrt(m) 1/m + e^m tan m + e^m sec^2 m ?
    Yes - see second 'spoiler' above.

    Quote Originally Posted by Hooperoo View Post
    sorry typed the question in wrong. the question is y + e^y = x-e^2x^2
    Right, so it's impicit differentiation using the chain rule...


    ... where (key in spoiler) ...

    Spoiler:


    ... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this case x), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).


    Spoiler:



    Now you may want to 'solve' for dy/dx, i.e. factorise the left hand side (of the bottom row) and then divide both sides by (1 + ey), to make dy/dx the 'subject'.

    _________________________________________

    Don't integrate - balloontegrate!

    Balloon Calculus; standard integrals, derivatives and methods

    Balloon Calculus Drawing with LaTeX and Asymptote!
    Last edited by tom@ballooncalculus; April 17th 2012 at 04:52 AM.
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  5. #5
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    Re: Differentation question

    so i should be going from

    {dy/dm} = 1/2 m^-{1/2} ln m + sqrt(m) {1/m} + e^{m} tan m + e^{m} sec^{2} m ?

    to

    {dy/dx} + e^{y} {dy/dx} = 1- e^{2x^{2}} 4x

    ?
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  6. #6
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    Re: Differentation question

    Sure... if, for some reason or other, you want to be going from

    the answer to the first problem

    to

    the answer to the second

    ?!
    Thanks from a tutor
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  7. #7
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    Re: Differentation question

    Im sorry sir this just confused me greatly and i didnt know what i was doing.

    ill try harder next time
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