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March 28, 2017

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dx/dy cos x sin y + x^5 = 11
i got dx/dy= (-cos x cos y)/ (-sin x sin y +5x^4)

But the wolfram got
(1- cos x cos y)/(5x^4-sin y sin x), where is the 1 come from?

  • Math - ,

    so we are differentiating
    cosx siny + x^5 = 11 implicitly ?

    cosx(cosy) dy/dx + siny(-sinx) + 5x^4 = 0

    dy/dx = (sinx siny - 5x^3)/(cosx cosy)

    which agrees with my Wolfram result

    http://www.wolframalpha.com/input/?i=find+dy%2Fdx+for+%28cos%28x%29%29%28sin%28y%29%29+%2B+x%5E5+%3D+11

  • Math - ,

    just noticed you wanted dx/dy (strange)

    cosx siny + x^5 = 11
    cosx(cosy) + siny(-sinx) dx/dy = 5x^4 dx/dy = 0
    dx/dy (5x4 - sinxsiny) = - cosxcosy
    dx/dy = -cosxcosy/(5x^4 - sinxsiny)

    the same as yours.

  • Math - ,

    At reiny .... Thanks

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