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Math Help - derivatives of trigo identity

  1. #1
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    Question derivatives of trigo identity

    I don't how to start with this kind of problem. Can someone show me the solution for this one.

    Find the first derivative

    sin^2(sin(sin x))
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  2. #2
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    Quote Originally Posted by jasonlewiz View Post
    I don't how to start with this kind of problem. Can someone show me the solution for this one.

    Find the first derivative

    sin^2(sin(sin x))
    u=sinx

    v=sinu

    w=sin^2v


    \displaystyle\frac{dw}{dx}=\frac{dw}{dv}\ \frac{dv}{du}\ \frac{du}{dx}

    =\displaystyle\frac{d}{dv}sin^2v\ \frac{d}{du}sinu\ \frac{d}{dx}sinx
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  3. #3
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    If you like you can write....

    u=Sinx

    v=Sinu

    w=Sinv

    t=w^2

    then, using the Chain Rule of differentiation...

    \displaystyle\frac{dt}{dx}=\frac{dt}{dw}\ \frac{dw}{dv}\ \frac{dv}{du}\ \frac{du}{dx}

    =\displaystyle\frac{d}{dw}w^2\ \frac{d}{dv}Sinv\ \frac{d}{du}Sinu\ \frac{d}{dx}Sinx

    =[2w][Cosv][Cosu][Cosx]=[2Sinv][Cos(Sinu)][Cos(Sinx)][Cosx]

    =[2Sin(Sinu)][Cos(Sin(Sinx))][Cos(Sinx)][Cosx]

    =[2Sin(Sin(Sinx))][Cos(Sin(Sinx))][Cos(Sinx)][Cosx]
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  4. #4
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    Just in case a picture helps...



    ... 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, initially x, then u, then later v), 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:


    Alternatively...


    _________________________________________

    Don't integrate - balloontegrate!

    Balloon Calculus; standard integrals, derivatives and methods

    Balloon Calculus Drawing with LaTeX and Asymptote!
    Last edited by tom@ballooncalculus; August 24th 2010 at 06:45 AM. Reason: alternative
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