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Thread: Calculus: Implicit Differentiation

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  1. #1
    Ciddieless since 2004
    Join Date
    Sep 2004
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    Ireland
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    Assuming that what I've done is not complete and utter bollocks...

    y^2 = (x^2 - 9) / (x^2 + 9)

    y = [(x^2 - 9) / (x^2 + 9)]^1/2 - (Square root of the right hand side)

    So, using the chain rule and the quotient rule:

    dy/dx = [1/2] . [[(x^2 - 9) / (x^2 + 9)]^-1/2] . [36x / (x^4+18x^2+81)]

    Or, see attachment.

    All that you need to realise that if y^2 = something, then y = the squared root of something. After that, it all falls into place and becomes just your everyday chain rule calculus.
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