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Differentiation of functions defined implicitly!? (AQA C4) watch

    • Thread Starter

    Can someone explain how to do this for me? I missed one lesson and a tonne of confusion then ensued! I've gone through it in my textbook like I do with everything else, but I still can't make real sense of it

    Thanks peoples

    Let's say we have the implicit equation xy^2 = 0, and we want to differentiate with respect to x (to find 'horizontal' stationary points, say). Then because we have an equality, we can differentiate both sides:

    \dfrac{d}{dx} xy^2 = \dfrac{d}{dx} 0

    Using the product rule on the LHS, we get that

    \dfrac{d}{dx} xy^2 = y^2\dfrac{dx}{dx} +x \dfrac{dy^2}{dx}

    The most important thing about implicit differentiation is that we can regard y as a function of x and use the chain rule on the 2nd term:

    \dfrac{dy^2}{dx} = \dfrac{dy^2}{dy} * \dfrac{dy}{dx}

    We can differentiate the first term (to get 2y) and then we just leave the second term as it is. Then, we get something which looks like y^2 + 2xy\frac{dy}{dx} = 0 and we can solve for dy/dx: \frac{dy}{dx} = \frac{-y}{2x}
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Updated: March 20, 2011
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