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Thread: Implicit Differentiation

  1. #1
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    Implicit Differentiation

    I'm currently studying maths, and need help with the following question:

    $\displaystyle x^4 + 4yx - 2y^5 = 3$

    I'm new to implicit differentiation, and have followed the book closely on this one (using dy/dx notation, which I am also new to). I've been using matlab to verify my answers, but it hasn't matched with this one.

    My answer $\displaystyle dy/dx = \frac{(4x^3 + 4y + 4x)}{-10y^4}$
    Matlab answer = $\displaystyle 4x^3 + 4y$

    Can someone please help me with this?
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  2. #2
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    When you want an error spotted in your work, it's best to show what you did step by step. Otherwise I have no idea where you made a mistake and the only way to help you is to type out a full solution myself, which is time consuming.

    Can you post your work?
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  3. #3
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    That makes sense... working out step by step below:

    Step One
    $\displaystyle \frac{d}{dx}(x^4+4xy-2y^5) = \frac{d}{dx}(3)$


    Next step: Calculating the derivative bit by bit

    $\displaystyle \frac{d}{dx}(x^4) + \frac{d}{dx}(4xy) - \frac{d}{dx}(2y^5) = 0$


    Now...finding derivatives (including product rule regarding d/dx(4xy))

    $\displaystyle 4x^3 + \frac{d}{dx}(4x)y + 4x\frac{d}{dx}(y) - \frac{dy}{dx}\frac{d}{dy}(2y^5) = 0$


    this is where I think I get stuck...most likely regarding product rule and dy/dx style notation (what is the difference between $\displaystyle \frac{d}{dx} \frac{dy}{dx} \frac{d}{dy}$ and $\displaystyle \frac{dx}{dy}$?

    $\displaystyle 4x^3 + 4y + 4x\frac{d}{dx}(y) - \frac{dy}{dx}10y^4 = 0$

    $\displaystyle 4x^3 + 4y + 4x -\frac{dy}{dx}10x^4 = 0$

    $\displaystyle \frac{dy}{dx} = \frac{4x^3 + 4y + 4x}{-10y^4}$

    Also, when using dy/dx notation I've noticed my uni-written coursebook uses brackets after dy/dx to denote un-derived components of the equation, then once they've been differentiated the brackets are removed to signify so. Am I right in saying this? (is that the rule of thumb?)
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