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Math Help - Differentiation

  1. #1
    Super Member Showcase_22's Avatar
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    Differentiation

    I am a 16 year old student who is due to start college. I need help with this question!

    Find an expression in terms of x and y for dy/dx, given that:

    y^3+(3x^2)y-4x=0

    Any help would be greatly appreciated.
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  2. #2
    TD!
    TD! is offline
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    If it's hard to solve to the form y = f(x), you can use implicit differentiation and solve for y'(x).
    In general, this will depend on both x and y, but that's allowed here

    To check your answer, I find:

    <br />
y' = \frac{{dy}}{{dx}} =  - \frac{{2\left( {3xy - 2} \right)}}{{3\left( {x^2  + y^2 } \right)}}<br />
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  3. #3
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    Hello, Showcase_22!

    This requires Implicit Differentiation.
    I hope you're familiar with it.


    Find an expression in terms of x and y for \frac{dy}{dx}
    given that: . y^3 + 3x^2y - 4x\:=\:0

    We have: . y^3 \quad+ \quad3x^2y \;\;- \;4x\;=\;0
    . . . . . . . . \downarrow . . . . . \downarrow . . . . . \downarrow . . \downarrow
    Then: . 3y^2\frac{dy}{dx} + \overbrace{3x^2\frac{dy}{dx} + 6xy} - 4 \;= \;0

    . . . . . Chain Rule . . Product Rule


    Now solve for \frac{dy}{dx} . . .

    We have: . 3y^2\frac{dy}{dx} + 3x^2\frac{dy}{dx} \:=\:4 - 6xy

    Factor: . (3y^2 + 3x^2)\frac{dy}{dx}\:=\:4-6xy

    Therefore: . \frac{dy}{dx}\:=\:\frac{4-6xy}{3y^2 + 3x^2}

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  4. #4
    Super Member Showcase_22's Avatar
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    How do you differentiate like that?

    this is what i'm doing:

    y^3+ (3x^2)y-4x=0

    3y^2(dy/dx)+(3x^2)(2x(dy/dx))+3y-4=0

    3y^2(dy/dx)+(3x^2)(2x(dy/dx))=4-3y

    (dy/dx)((3y^2)+6x^3)=4-3y

    dy/dx=(4-3y)/((3y^2)+6x^3)

    It looks a little confusing like this, but the answer I get is a little wrong. How would you solve this?
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  5. #5
    Super Member Showcase_22's Avatar
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    Wow, thanks for your reply!

    I worked on that question for an hour this morning whilst on the bus, I thought it was impossible!
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