# How do you find dy/dx by implicit differentiation given 2x^2-3y^2=4?

Feb 10, 2017

Take the derivative of both sides with respect to $x$, treating $y$ as a function of $x$. Solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$.

#### Explanation:

$2 {x}^{2} - 3 {y}^{2} = 4$

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

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

$\implies \text{ "4x" "-" "6y dy/dx" "=0" }$(by chain rule)

$\implies \text{ "-6y dy/dx" } = - 4 x$

$\implies \text{ "dy/dx" "=" "("-"4x)/("-"6y)" "=" } \frac{2 x}{3 y}$

Therefore, $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x}{3 y}$.