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 Question from Katie, a student: find y' of x^2y-2y^3=3x+2y

Hi Katie,

Let me try this similar problem

Find $y^{\prime}$ if $x^2 y^2 + 2xy = x^2 - 2y.$

You are correct in thinking that this is an implicit differentiation problem. You are expected to see $x$ as the variable and $y$ as a function of $x,$ $y = y(x).$ You are then to find $y^{\prime} = \large \frac{dy}{dx}.$

The expression $x^2 y^2$ is a product of powers and you are going to have to use the product rule.

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

For each of the two derivatives you now need to use the power rule but since $y$ is a function of $x$ and you are differentiating with respect to $x$ you will also need to use the chain rule to differentiate $y^2.$ Hence

$\frac{d}{dx}\left(x^2 y^2\right) = x^2 \left( 2 y \frac{dy}{dx} \right) + y^2 (2 x) = 2 x^2 y \frac{dy}{dx} + 2 x y^2$

The remaining three terms are easier so when I differentiate both sides of $x^2 y^2 + 2xy = x^2 - 2y$ I get

$2 x^2 y y^{\prime} + 2 x y^2 + 2 x y^{\prime} + 2 y = 2 x - 2 y^{\prime}$

Notice that there is a common factor of 2 on both sides so I divide both sides by 2 and collect all the terms involving $y^{\prime}$ on the left side and the remaining terms on the right side. This gives me

$\left(x^2 y + x + 1\right) y^{\prime} = x - y - x y^2$

and hence

$y^{\prime} = \frac{x - y - x y^2}{x^2 y + x + 1}$