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Math Help - Implicit Differentiation....

  1. #1
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    Implicit Differentiation....

    This should be fairly simple: x^2+y=x^3+y^3

    Need help understanding how to work it out and what (d/dx) and (dx/dy) mean and how they help solve the problem.
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  2. #2
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    When you differentiate y, stick a dy/dx beside it and then solve for dy/dx.

    x^{2}+y-x^{3}-y^{3}=0

    2x+\frac{dy}{dx}-3x^{2}-3y^{2}\frac{dy}{dx}=0

    Solve for dy/dx.
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  3. #3
    Super Member angel.white's Avatar
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    I would like to point out that \frac{dy}{dx} means the derivative of y with respect to x. In actuality, you solve both x and y in the exact same manner, it just doesn't look this way because \frac{dx}{dx} = 1 which is usually an invisible coefficient.

    For example, the derivative of y=x is
    \frac{dy}{dx} = \frac{dx}{dx}
    and \frac{dx}{dx} = 1, so \frac{dy}{dx} = 1

    Another way of saying \frac{dy}{dx} would be y prime, or y\prime.

    ------

    So for your problem:
    x^{2}+y=x^{3}+y^{3}

    Prep - First, you need to know the Power Rule and the Chain Rule, the Power Rule says that the derivative of x^{n} with respect to x is equal to nx^{n-1}. So for example, \frac{d}{dx} x^{5} = 5x^{4} This can be proven, but that's beyond the scope of this post. Don't get confused by \frac{d}{dx} It just means the derivative of ... with respect to x, where ... is whatever it is being multiplied by, in this case x^{5}.
    The Chain Rule says that the derivative of f(g(x)) is equal to f\prime(g(x))g\prime(x). Remember that f\prime(x) = \frac{d}{dx}f(x) = \frac{df(x)}{dx}. So if f(x)=x^{5}, and g(x)=x+1, then f(g(x))=(x+1)^{5}. Now differentiate, \frac{d}{dx}f(g(x))=f\prime(g(x))g(x) So, just fill in the equation, \frac{d}{dx}(x+1)^{5} = 5(x+1)^{4}(1), because the derivative of g(x) is 1.

    Step1 - differentiate both sides:
    2x\frac{dx}{dx} + \frac{dy}{dx} = 3x^{2}\frac{dx}{dx} + 3y^{2}\frac{dy}{dx}

    This entire step is Power Rule combined with Chain Rule, y^3 can be looked at as 2 functions, f(x) = x^{3} and g(x) = y. So f(g(x)) = y^{3}. This means f\prime(g(x)) = 3y^{2} and g\prime(x) = dy/dx or y\prime

    Step2 - Simplify dx/dx as 1
    2x + y\prime = 3x^{2} + 3y^{2}y\prime

    step3 - Combine like terms (put any y\prime on the same side, and anything else on the other side)
    y\prime - 3y^{2}y\prime = 3x^{2} - 2x

    step4 - Factor out y\prime
    y\prime(1 - 3y^{2}) = 3x^{2} - 2x

    step5 - Solve for y\prime in this case, it means divide out everything it is being multiplied by.
    y\prime = \frac{3x^{2} - 2x}{1 - 3y^{2}}


    So anyway, unless I messed up somewhere, that's the correct process for finding the answer for this type of a question.
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