# Thread: Another Implicit Diff Problem

1. ## Another Implicit Diff Problem

Solving for y'

$\displaystyle f(x) = 8x^{2} + 3xy - y^{2} = 2$

$\displaystyle 16x + (y)(3)y' + (3x)(1)y' - 2yy' = 0$ - Product Rule on $\displaystyle 3xy$

$\displaystyle 16x + (3)(y)y' + (3x)(1)y' - 2yy' = 0$

$\displaystyle 16x + 3yy' + 3xy' - 2yy' = 0$

$\displaystyle 16x + y'(3y + 3x - 2y) = 0$

$\displaystyle y'(3y + 3x - 2y) = -16x$

$\displaystyle y' = \dfrac{16x}{3y + 3x - 2y}$

2. ## Re: Another Implicit Diff Problem

Hello, Jason76!

$\displaystyle f(x) \:=\:8x^2 + 3xy - y^2 \:=\:0\quad \text{ Find }y'.$

We have: . . . .$\displaystyle 8x^2 + 3xy - y^2 \:=\:0$

Then: .$\displaystyle 16x + \overbrace{3xy' + 3y} - 2yy' \:=\:0$

n . . . . . . . . . . . . $\displaystyle 3xy' - 2yy' \:=\:-16x - 3y$

n . . . . . . . . . . . . $\displaystyle (3x-2y)y' \:=\:-(16x+3y)$

. . . . . . . . . . . . . . . . . . . . $\displaystyle y' \:=\:-\frac{16x+3y}{3x-2y}$

3. ## Re: Another Implicit Diff Problem

Right??

$\displaystyle f(x) = 8x^{2} + 3xy - y^{2} = 2$

$\displaystyle 16x + (y)(3) + (3x)(1)y' - 2yy' = 0$ - Product Rule on $\displaystyle 3xy$

$\displaystyle 16x + (3)(y) + (3x)(1)y' - 2yy' = 0$

$\displaystyle 16x + 3y + 3xy' - 2yy' = 0$

$\displaystyle 16x + 3y + y'(3x - 2y) = 0$

$\displaystyle y'(3x - 2y) = -16x - 3y$

$\displaystyle y'= \dfrac{-16x - 3y}{3y - 2y}$