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

Question

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

Luke Alvarez · Accepted Answer

#dy/dx =- (2x^2y + 3xy^2)/(x^3 + 2x^2y - 6)#

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

#x^3y + x^2y^2 - 5y - y = 0#

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

#3x^2y + x^3(dy/dx) + 2xy^2 + 2yx^2(dy/dx) - 6(dy/dx) = 0#

#x^3(dy/dx) + 2x^2y(dy/dx) - 6(dy/dx) = -3x^2y - 2xy^2#

#dy/dx(x^3 + 2x^2y - 6) = -3x^2y - 2xy^2#

#dy/dx =- (2x^2y + 3xy^2)/(x^3 + 2x^2y - 6)#

Hopefully this helps!

Ezra Adams · Answer

undefined To find the implicit derivative of $ y = x^3y + x^2y^4 - 5y $, differentiate both sides of the equation with respect to $ x $, treating $ y $ as a function of $ x $ using the product rule and chain rule as needed.

$$
\frac{{dy}}{{dx}} = \frac{{d}}{{dx}}(x^3y) + \frac{{d}}{{dx}}(x^2y^4) - \frac{{d}}{{dx}}(5y)
$$

Apply the product rule and chain rule as necessary:

$$
\frac{{dy}}{{dx}} = 3x^2y + x^3\frac{{dy}}{{dx}} + 2xy^4 + x^2(4y^3)\frac{{dy}}{{dx}} - 5\frac{{dy}}{{dx}}
$$

Rearrange terms to isolate $ \frac{{dy}}{{dx}} $:

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

Factor out $ \frac{{dy}}{{dx}} $:

$$
\left(1 - x^3 - 4x^2y^4\right)\frac{{dy}}{{dx}} = 3x^2y + 2xy^4 - 5y
$$

Divide both sides by $ 1 - x^3 - 4x^2y^4 $ to solve for $ \frac{{dy}}{{dx}} $:

$$
\frac{{dy}}{{dx}} = \frac{{3x^2y + 2xy^4 - 5y}}{{1 - x^3 - 4x^2y^4}}
$$