# How do you implicitly differentiate -3=5x^3y-x^2y+y^2/x?

Nov 29, 2015

y'=$\frac{{y}^{2} + 2 {x}^{3} y - 15 {x}^{4} y}{5 {x}^{5} - {x}^{4} + 2 x y}$

#### Explanation:

$5 {x}^{3} y - {x}^{2} y + {y}^{2} / x$=-3
Differentiating on both sides with respect to x

d/dx( 5x^3y)-d/dx(-x^2y)+d/dx(y^2/x)=d/dx(-3)

Use product rule for first two and quotient rule for third part

$15 {x}^{2} y + 5 {x}^{3} y ' - 2 x y - {x}^{2} y ' + \frac{2 y y ' x - {y}^{2}}{x} ^ 2$=0

$\frac{15 {x}^{4} y + 5 {x}^{5} y ' - 2 {x}^{3} y - {x}^{4} y ' + 2 y y ' x - {y}^{2}}{x} ^ 2$=0

A rational expression is 0 , only if the numerator is 0

so $\left(15 {x}^{4} y + 5 {x}^{5} y ' - 2 {x}^{3} y - {x}^{4} y ' + 2 y y ' x - {y}^{2}\right)$=0

solve for y'

$\left(5 {x}^{5} - {x}^{4} + 2 x y\right) y ' = {y}^{2} + 2 {x}^{3} y - 15 {x}^{4} y$

y'=$\frac{{y}^{2} + 2 {x}^{3} y - 15 {x}^{4} y}{5 {x}^{5} - {x}^{4} + 2 x y}$