2
$\begingroup$

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.

  • 0
    @julianKueshammer Did you find any mistakes?2012-11-14
  • 0
    It's not the same as $E$, it has a different sign in the denominator. However, it would be the same as $E$ if you hadn't switched that sign along the way.2012-11-14
  • 0
    @joriki, Where did I switch the sign?2012-11-14
  • 0
    Going from the first to the second line you did some silly error2012-11-14
  • 0
    @MaoYiyi I just made the title more descriptive.2012-11-14
  • 0
    @ Mao Yiyi: Now it's okay2012-11-14
  • 0
    How could **A** be the answer?2012-11-14
  • 0
    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

2

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, $$y'=\frac{15x^2+4y}{-4(x+y)}$$

  • 0
    F(x,y) = 1, not zero. ??? COuld you explain a bit more?2012-11-14
  • 0
    How does one get that formula. I tried searching for the LaTeX of your answer, but I didn't find anything that would explain it.2012-11-14
  • 1
    @MaoYiyi: Let $F(x,y)=0$. Now take the differential of it, considering that $y$ is a function of $x$. Then you get $$F_x(x,y)\frac{dx}{dx}+F_y(x,y)\frac{dy}{dx}=0$$ Rearranging the last equation, we have above.2012-11-15
  • 0
    @amWhy: More than expected are yours now Amy. Thanks dear.2013-04-04
0

$$-5x^3-4xy-2y^2 = 1$$ $$-15x^2-4y-4xy'-4yy'=0$$ $$4xy'+4yy'=-15x^2-4y$$ $$y'(4x+4y)=-15x^2-4y$$ $$y'=\frac{-15x^2-4y}{4x+4y}$$

  • 1
    This is what just Mao has given , it's nothing new!2012-11-14
  • 1
    that is right answer2012-11-14