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Math Help - Implicit differentiation

  1. #1
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    Implicit differentiation

    1.x^2+3xy+y^2=15 And the answer says -y^2/x^2
    2.(1/x)+(1/y) the answer says -y^2/x^2.

    For 1., dy/dx(x^2+3xy+y^2)=dy/dx(15)->(2x+3x+3y(dy/dx)+2y(dy/dx)=0 (for 3xy, I used the product rule for x and y and got x+y and plugged it to 3 to get this function.) What am i doing wrong with this equation?

    With 2. I used the quotient rule, and got (x/x^2)+(y/y^2) and I know I am wrong on this one from the start. please help.
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  2. #2
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    For #1:

    x^{2}+3xy+y^{2}=15

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

    \frac{dy}{dx}=\frac{-(3y+2x)}{3x+2y}

    Are you sure that's the book answer?. It isn't \frac{-y^{2}}{x^{2}}
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  3. #3
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    Quote Originally Posted by driver327 View Post
    1.x^2+3xy+y^2=15 And the answer says -y^2/x^2
    2.(1/x)+(1/y) the answer says -y^2/x^2.

    For 1., dy/dx(x^2+3xy+y^2)=dy/dx(15)->(2x+3x+3y(dy/dx)+2y(dy/dx)=0 (for 3xy, I used the product rule for x and y and got x+y and plugged it to 3 to get this function.) What am i doing wrong with this equation?

    With 2. I used the quotient rule, and got (x/x^2)+(y/y^2) and I know I am wrong on this one from the start. please help.
    Hello,

    could it be that the first answer is missing?

    To derivate implicitely I use the formula:

    If f(x,y) = 0 then \frac{df}{dx}=y'=-\frac{\frac{\partial{f}}{\partial{x}}}{\frac{\part  ial{f}}{\partial{y}}}

    to #1:
    f(x,y) = x^2+3xy+y^2-15=0

    y' = -\frac{2x+3y}{3x+2y}

    to #2:

    Using the same method. f(x,y) = \frac{1}{x} + \frac{1}{y} = \frac{x+y}{x\cdot y} then (using quotient rule!)

    y' = -\frac{\frac{xy - (x+y)y}{x^2 \cdot y^2}}{\frac{xy-(x+y)x}{x^2 \cdot  y^2}} = -\frac{-y^2}{-x^2} = -\frac{y^2}{x^2}
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  4. #4
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    I accidentally put down the same answer. But the first one is correct.
    Last edited by driver327; July 20th 2007 at 05:06 PM.
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