# How do you implicitly differentiate -4=(2x+3y)^2-y^2x ?

$y ' = \frac{{y}^{2} - 12 y - 8 x}{2 \left(6 x + 9 y - x y\right)}$

#### Explanation:

We start from the given equation and differentiate both sides of the equation with respect to x

$- 4 = {\left(2 x + 3 y\right)}^{2} - {y}^{2} x$

$\frac{d}{\mathrm{dx}} \left(- 4\right) = \frac{d}{\mathrm{dx}} \left({\left(2 x + 3 y\right)}^{2} - {y}^{2} x\right)$

$0 = 2 {\left(2 x + 3 y\right)}^{2 - 1} \cdot \frac{d}{\mathrm{dx}} \left(2 x + 3 y\right) - \frac{d}{\mathrm{dx}} \left({y}^{2} x\right)$

0=2(2x+3y)*(2+3y')-[y^2*1+x*2y*y'])

Simplify at this point to obtain the $y '$

$0 = \left(4 x + 6 y\right) \cdot \left(2 + 3 y '\right) - {y}^{2} - 2 x y y '$

$0 = \left(8 x + 12 y + 12 x y ' + 18 y y '\right) - {y}^{2} - 2 x y y '$

$0 = 8 x + 12 y + 12 x y ' + 18 y y ' - {y}^{2} - 2 x y y '$

$\left({y}^{2} - 12 y - 8 x\right) = \left(12 x + 18 y - 2 x y\right) y '$

$y ' = \frac{{y}^{2} - 12 y - 8 x}{12 x + 18 y - 2 x y}$

$y ' = \frac{{y}^{2} - 12 y - 8 x}{2 \left(6 x + 9 y - x y\right)}$

God bless....I hope the explanation is useful.