# How do you implicitly differentiate  x^3 + y/x^2 + y^3 = 8xy?

Sep 27, 2016

Remove each term and differentiate, put each derivative back in equation, move the terms containing $\frac{\mathrm{dy}}{\mathrm{dx}}$ to the left and the others to the right, divide both sides by the coefficient of $\frac{\mathrm{dy}}{\mathrm{dx}}$

#### Explanation:

Term 1:

$\frac{d \left[{x}^{3}\right]}{\mathrm{dx}} = 3 {x}^{2}$

Term 2:

Use the quotient rule, $\frac{d \left[\left(\frac{g}{h}\right)\right]}{\mathrm{dx}} = \frac{g ' h - h ' g}{h} ^ 2$

let $g = y$, then $g ' = \frac{\mathrm{dy}}{\mathrm{dx}}$, $h = {x}^{2}$, $h ' = 2 x$ and ${h}^{2} = {x}^{4}$

$\frac{d \left[\left(\frac{y}{{x}^{2}}\right)\right]}{\mathrm{dx}} = \frac{{x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} - 2 x y}{x} ^ 4$

$\frac{d \left[\left(\frac{y}{{x}^{2}}\right)\right]}{\mathrm{dx}} = \frac{1}{x} ^ 2 \frac{\mathrm{dy}}{\mathrm{dx}} - 2 \frac{y}{x} ^ 3$

Term 3:

$\frac{d \left[{y}^{3}\right]}{\mathrm{dx}} = 3 {y}^{2} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)$

Term 4:
Use the product rule $\frac{d \left[g h\right]}{\mathrm{dx}} = g ' h + g h '$

let $g = 8 x$, then $h = y$, $g ' = 8$, and $h ' = \frac{\mathrm{dy}}{\mathrm{dx}}$

$\frac{d \left[8 x y\right]}{\mathrm{dx}} = 8 y + 8 x \frac{\mathrm{dy}}{\mathrm{dx}}$

Put the derivatives back into their corresponding locations in the equation:

$3 {x}^{2} + \frac{1}{x} ^ 2 \frac{\mathrm{dy}}{\mathrm{dx}} - 2 \frac{y}{x} ^ 3 + 3 {y}^{2} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 8 y + 8 x \frac{\mathrm{dy}}{\mathrm{dx}}$

Move the terms containing $\frac{\mathrm{dy}}{\mathrm{dx}}$ to the left and the others to the right:

$\frac{1}{x} ^ 2 \frac{\mathrm{dy}}{\mathrm{dx}} + 3 {y}^{2} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) - 8 x \frac{\mathrm{dy}}{\mathrm{dx}} = 8 y + 2 \frac{y}{x} ^ 3 - 3 {x}^{2}$

Factor $\frac{\mathrm{dy}}{\mathrm{dx}}$ from the left:

$\left\{\frac{1}{x} ^ 2 + 3 {y}^{2} - 8 x\right\} \frac{\mathrm{dy}}{\mathrm{dx}} = 8 y + 2 \frac{y}{x} ^ 3 - 3 {x}^{2}$

Multiply both sides by ${x}^{3}$:

$\left\{x + 3 {y}^{2} {x}^{3} - 8 {x}^{4}\right\} \frac{\mathrm{dy}}{\mathrm{dx}} = 8 y {x}^{3} + 2 y - 3 {x}^{5}$

Divide both sides by the coefficient of $\frac{\mathrm{dy}}{\mathrm{dx}}$:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{8 y {x}^{3} + 2 y - 3 {x}^{5}}{x + 3 {y}^{2} {x}^{3} - 8 {x}^{4}}$