# How do you implicitly differentiate  xy + 3x + 5x^2 = 4/y^3-xy+1?

$y ' = \frac{- 10 x - 2 y - 3}{2 x + \frac{12}{y} ^ 4}$

#### Explanation:

The given equation

$x y + 3 x + 5 {x}^{2} = \frac{4}{y} ^ 3 - x y + 1$

Differentiate both sides of the equation with respect to x

$\frac{d}{\mathrm{dx}} \left(x y + 3 x + 5 {x}^{2}\right) = \frac{d}{\mathrm{dx}} \left(\frac{4}{y} ^ 3 - x y + 1\right)$

$\frac{d}{\mathrm{dx}} \left(x y\right) + \frac{d}{\mathrm{dx}} \left(3 x\right) + \frac{d}{\mathrm{dx}} \left(5 {x}^{2}\right) = \frac{d}{\mathrm{dx}} \left(\frac{4}{y} ^ 3\right) - \frac{d}{\mathrm{dx}} \left(x y\right) + \frac{d}{\mathrm{dx}} \left(1\right)$

$x y ' + y \left(1\right) + 3 + 10 x = 4 \left(- 3\right) {y}^{- 3 - 1} \cdot y ' - \left[x y ' + y \cdot 1\right] + 0$

simplify

$x y ' + y + 3 + 10 x = - 12 {y}^{- 4} \cdot y ' - x y ' - y$

transpose all terms containing y' on one side of the equation

$x y ' + 12 {y}^{- 4} y ' + x y ' = - y - y - 10 x - 3$

$2 x y ' + 12 {y}^{- 4} y ' = - 2 y - 10 x - 3$

$\left(2 x + 12 {y}^{- 4}\right) y ' = - 2 y - 10 x - 3$

$y ' = \frac{- 10 x - 2 y - 3}{2 x + \frac{12}{y} ^ 4}$

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