How do you find (dy)/(dx) given -3x^2y^2-2y^3+5=5x^2?

Sep 22, 2017

(dy)/(dx)=(-(5x+3xy^2))/(3y(x^2+y)

Explanation:

we want

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

We will differentiate implicitly.

The first term will also need the product rule

$\textcolor{red}{\frac{d}{\mathrm{dx}} \left(u v\right) = v \frac{\mathrm{du}}{\mathrm{dx}} + u \frac{\mathrm{dv}}{\mathrm{dx}}}$

${y}^{2} \left(- 6 x\right) + \left(- 3 {x}^{2}\right) 2 y \frac{\mathrm{dy}}{\mathrm{dx}} - 6 {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} + 0 = 10 x$

$- 6 x {y}^{2} - 6 {x}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}} - 6 {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} = 10 x$

now rearrange for $\frac{\mathrm{dy}}{\mathrm{dx}}$ and tidy up.

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(- 6 {x}^{2} y - 6 {y}^{2}\right) = 10 x + 6 x {y}^{2}$

(dy)/(dx)=(10x+6xy^2)/(-6y(x^2+y)

(dy)/(dx)=(-(5x+3xy^2))/(3y(x^2+y)