Thread: [SOLVED] Implicit Differentiation

x^3 * y^3 - y = x

help please

differentiate implicitly with respect to x...

Originally Posted by edge

x^3 * y^3 - y = x

help please

$\displaystyle x^3 \cdot 3y^2 \frac{dy}{dx} + y^3 \cdot 3x^2 - \frac{dy}{dx}=1$

remember to use product rule on the first.

Okay, thanks... now just to make sure I'm doing this right, how would you solve for dy/dx?

Originally Posted by GaloisTheory1

$\displaystyle x^3 \cdot 3y^2 \frac{dy}{dx} + y^3 \cdot 3x^2 - \frac{dy}{dx}=1$

remember to use product rule on the first.

$\displaystyle x^3 \cdot 3y^2 \frac{dy}{dx} + y^3 \cdot 3x^2 - \frac{dy}{dx}=1$

so

$\displaystyle \frac{dy}{dx}=\frac{1-y^3 \cdot 3x^2}{x^3 \cdot 3y^2-1}$