# How do you implicitly differentiate 1=x^3y + 5xy ?

Dec 25, 2015

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{3 {x}^{2} y + 5 y}{{x}^{3} + 5 x}$

#### Explanation:

Use the product rule in both terms on the right hand side.

Recall that implicitly differentiating a $y$ term will create a $\frac{\mathrm{dy}}{\mathrm{dx}}$ term.

$0 = 3 {x}^{2} y + {x}^{3} \frac{\mathrm{dy}}{\mathrm{dx}} + 5 y + 5 x \frac{\mathrm{dy}}{\mathrm{dx}}$

Isolate the $\frac{\mathrm{dy}}{\mathrm{dx}}$ terms to solve for them.

$- 3 {x}^{2} y - 5 y = \frac{\mathrm{dy}}{\mathrm{dx}} \left({x}^{3} + 5 x\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{3 {x}^{2} y + 5 y}{{x}^{3} + 5 x}$