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Math Help - Implicit Differentiation

  1. #1
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    Implicit Differentiation

    I got past the first step, but I am having trouble solving for d/dx

    (x^(2) + y^(2))^(3) = ax^(2)y a is constant

    for the first step i got

    3(x^(2) + y^(2))^(2) * (2x + 2y (d/dx)) = ax^(2) (d/dx) + 2axy


    after that I'm stuck...
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by drain View Post
    I got past the first step, but I am having trouble solving for d/dx

    (x^(2) + y^(2))^(3) = ax^(2)y a is constant

    for the first step i got

    3(x^(2) + y^(2))^(2) * (2x + 2y (d/dx)) = ax^(2) (d/dx) + 2axy


    after that I'm stuck...
    dont you mean you want to solve for dy/dx?

    (x^(2) + y^(2))^(3) = ax^(2)y ..........its painful, but i think it'll help to expand these brackets.

    => x^6 + 3x^4y^2 + 3x^2y^4 + y^6 = ax^2y

    now we go through and differentiate implicitly. we add dy/dx to a function when we differentiate with respect to y, we add dx/dx when we differentiate with respect to x, but dx/dx cancels into 1, so we actually don't write it. also note that we need the product rule when we have x * y in any form. so here goes.

    => 6x^5 + 12x^3y^2 + 6x^4y dy/dx + 6xy^4 + 12x^2y^3 dy/dx + 6y^5 dy/dx = 2axy + ax^2 dy/dx ...........now we group everything with dy/dx on one side so we can factor them out.

    => 6x^4y dy/dx + 12x^2y^3 dy/dx + 6y^5 dy/dx - ax^2 dy/dx = 2axy - 6x^5 - 12x^3y^2 - 6xy^4

    => dy/dx(6x^4y + 12x^2y^3 + 6y^5 - ax^2) = 2axy - 6x^5 - 12x^3y^2 - 6xy^4
    => dy/dx = (2axy - 6x^5 - 12x^3y^2 - 6xy^4)/(6x^4y + 12x^2y^3 + 6y^5 - ax^2)


    check if i expanded the cubic correctly
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  3. #3
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    Hello, drain!

    I got past the first step, but I am having trouble solving for d/dx

    . . (x + y) .= .axy . . a is constant

    For the first step i got:
    . . 3(x + y)(2x + 2yy') .= .axy' + 2axy . Good!

    Expand the left side (partially): .6x(x + y) + 6y(x + y)y' .= .axy' + 2axy

    Rearrange terms: .6y(x + y)y' - axy' .= .2axy - 6x(x + y)

    Factor: . [6y(x + y) - ax]y' .= .2x[ay - 3(x + y)]

    . . . . . . . . . . . . . . 2x[ay - 3(x + y)]
    Therefore: . y' . = . ------------------------
    . . . . . . . . . . . . . . .6y(x + y) - ax

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  4. #4
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    Thanks alot! That one was realllly tricky... and the way that other user did it... I appreciate the help but it confused the heck out of me!
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