# Math Help - Another Implicit Diff Problem

1. ## Another Implicit Diff Problem

Solving for y'

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

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

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

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

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

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

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

2. ## Re: Another Implicit Diff Problem

Hello, Jason76!

Your "product rule" is incorrect.

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

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

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

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

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

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

3. ## Re: Another Implicit Diff Problem

Right??

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

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

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

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

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

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

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