What is the implicit derivative of y=x^3y+x^2y^4-5y ?

Aug 24, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{2 {x}^{2} y + 3 x {y}^{2}}{{x}^{3} + 2 {x}^{2} y - 6}$

Explanation:

Let's first put all our variables on one side.

${x}^{3} y + {x}^{2} {y}^{2} - 5 y - y = 0$

By a combination of implicit differentiation and the product rule, we can differentiate without having to isolate $y$.

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

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

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{2 {x}^{2} y + 3 x {y}^{2}}{{x}^{3} + 2 {x}^{2} y - 6}$

Hopefully this helps!