Question

If y3 + xy2-2x = 0 defines y implicitly as a function of x, then the value of dx dy at the point (4, -2) is (A) - 2 1 (B)- 8 1 (C) 4 1 (D) 2 1

Answer 1

im super hay - my h and put a g

Answer 2

Employ the chain rule just as you would normally to find \(dy/dx\). This should give you a start.\[\color{blue}{\frac{d}{dx}}\left(y^3 + \underline{xy^2}-2x \right)= \color{blue}{\frac{d}{dx}}0\]\[3y^2\color{blue}{\frac{dy}{dx}}+\underline{\left(y^2+2xy\color{blue}{\frac{dy}{dx}}\right)}-2=0\]

Answer 3

use implicit differentiation formula to find dx/dy:
\[dx/dy= - F _{y}/F _{x}\]
where F = y3 + xy2-2x = 0

Answer 4

d