How do you differentiate #-1=xy^3-x^2y#?

Question

Ezekiel Alexander · Accepted Answer

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

When doing implicit differentiation, if differentiating #y# with respect to #x# then #x# can be differentiated as usual (i.e. #d/dx(x) = 1#).

But #y# is a bit different. The derivative of #y# alone will not be #1# but #dy/dx#.

For any power or function of y the chain rule will have to be implemented. For example, the derivative of #y^3# will be #3y^2dy/dx#. To differentiate #-1=xy^3-x^2y# simply differentiate term by term. So:

#d/dx(-1)=0# #d/dx(xy^3)=y^3+3xy^2dy/dx# (Don't forget the product rule!) #d/dx(-x^2y) = -2xy-x^2dy/dx#

So putting these back into the given function; differentiating #-1=xy^3-x^2y# will give us:

#0=y^3+3xy^2dy/dx-2xy-x^2dy/dx#

Rearrange this to get dy/dx on one side of the equation:

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

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

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

And, thus our final solution.

Charles Arocha · Answer

undefined To differentiate $-1 = xy^3 - x^2y$:

1. Differentiate each term of the equation with respect to $x$.
2. Use the product rule and chain rule where necessary.
3. Collect terms involving $dy/dx$ on one side and terms involving $y$ and $x$ on the other side.
4. Solve for $dy/dx$.

The derivative of the given equation with respect to $x$ is:

$$y^3 + 3xy^2\frac{dy}{dx} - 2xy - x^2\frac{dy}{dx} = 0$$

Solve for $\frac{dy}{dx}$:

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