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From my last test.

Given that $-5x^3-4xy-2y^2 = 1$. Determine the change in $y$ with respect to $x$.

A. $-\dfrac{-15x^2-4}{-4-4y}$ B. $- \dfrac{-15x^2-4y}{-4-4y}$ C. $- \dfrac{-15x^2-4}{-4x-4y}$ D. $- \dfrac{-10x-4y}{-4x-2} $ E. $- \dfrac{-15x^2-4y}{-4x-4y}$

I got the answer E, but the teacher said it was A.

My Work (tell me my mistake)

find $dx$ and $dy$ at the same time.

$dx(-15x^2-4y)-dy(4x+4y)=0$

$dx(-15x^2-4y)=dy(4x+4y)$

$\dfrac{dx(-15x^2-4y)}{dy(4x+4y)} = 1$

$\left(\dfrac{dy}{dx}\right) \dfrac{dx(-15x^2-4y)}{dy(4x+4y)} = 1\left(\dfrac{dy}{dx}\right)$

$ \dfrac{-15x^2-4y}{4x+4y} = \dfrac{dy}{dx}$ which is the same as E.

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    I guess the simple solution to that question is that A is not the answer but$E$is. You can cross check that at: [Wolfram|Alpha $-5x^3+-4xy-2y^2=1$](http://www.wolframalpha.com/input/?i=-5x^3+-4x*y-2y^2=1) in section implicit derrivatives.2012-11-14

2 Answers 2

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If we have a relation $F(x,y)=0$ in which $y$ is a function of $x$ then : $y'=\frac{-F_x}{F_y}$ Here, $F(x,y)=-5x^3-4xy-2y^2-1=0$ and $F_x=-15x^2-4y, F_y=-4x-4y$. So, $y'=\frac{15x^2+4y}{-4(x+y)}$

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    @amWhy: More than expected are yours now Amy. Thanks dear.2013-04-04
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$-5x^3-4xy-2y^2 = 1$ $-15x^2-4y-4xy'-4yy'=0$ $4xy'+4yy'=-15x^2-4y$ $y'(4x+4y)=-15x^2-4y$ $y'=\frac{-15x^2-4y}{4x+4y}$

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    that is right answer2012-11-14