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.