Implicit Derivatives of Polynomial Relations

Why Polynomial Relations Need Care

A polynomial relation can mix powers of \(x\) and \(y\), even when it does not express \(y\) explicitly as a function of \(x\). For example, the relation \(xy+y^3=10\) describes points on a curve. At a point on that curve where \(y\) varies differentiably with \(x\), we can find the slope by differentiating both sides with respect to \(x\).

The important detail is that \(y\) depends on \(x\). A term involving \(y\) therefore contributes a factor of \(\frac{dy}{dx}\) when differentiated. A mixed term, such as \(x^2y\), has both an \(x\)-factor and a \(y\)-factor to account for.

How to Differentiate the Terms

The power rule differentiates a power of \(x\) in the usual way. If \(y=y(x)\), the chain rule gives a different-looking result for a power of \(y\):

For a mixed monomial, differentiate its factors while accounting for both their dependence on \(x\). For example, \(xy\) has one factor that is \(x\) and another factor that is \(y(x)\). Differentiating gives \(y+x\frac{dy}{dx}\). Similarly, \(x^2y\) gives \(2xy+x^2\frac{dy}{dx}\): the first term comes from differentiating \(x^2\), and the second from differentiating \(y\).

A useful way to keep the calculation organized is to write the derivative of each original term on its own line. In particular, do not omit a term just because it contains both \(x\) and \(y\). Each factor can contribute to the derivative.

Why the Mixed-Term Rule Works

The chain rule explains why a power of \(y\) contributes \(\frac{dy}{dx}\). The product rule explains how the \(x\)- and \(y\)-factors of a mixed monomial both contribute. For \(x^m y^n\), with \(y=y(x)\), apply the product rule to the two factors \(x^m\) and \(y^n\):

This derivation assumes \(m\) and \(n\) are nonnegative integers, as they are for polynomial terms, and that \(y\) is differentiable as a function of \(x\) along the curve being considered. It gives two contributions: one from differentiating the power of \(x\), and one from differentiating the power of \(y\). If either exponent is zero, its corresponding contribution is zero.

For an entire polynomial, apply this rule term by term and add the results. The original relation is constant, often zero, so its derivative is zero. The resulting equation can then be rearranged to isolate \(\frac{dy}{dx}\), provided its coefficient is nonzero.

A Reliable Procedure

Worked Examples

A Common Sign and Dependence Mistake

A frequent error is to differentiate \(x^2y\) as though \(y\) were constant, or to attach \(\frac{dy}{dx}\) to both contributions. For instance, the incorrect calculation

misses the contribution from differentiating \(y\) after the \(x^2\) factor is left unchanged, and it also incorrectly places \(\frac{dy}{dx}\) on the contribution from differentiating \(x^2\). The product rule gives the correct result:

A useful check is to ask which factor was differentiated in each term. The term \(2xy\) comes from differentiating \(x^2\), so it has no \(\frac{dy}{dx}\). The term \(x^2\frac{dy}{dx}\) comes from differentiating \(y\), so it does.

When Solving for the Derivative Is Not Possible

After differentiation and collection, the equation may have the form

If \(A(x,y)\ne 0\) at the point under consideration, algebra gives \(\frac{dy}{dx}=\frac{B(x,y)}{A(x,y)}\). If \(A(x,y)=0\), this division is not valid. The differentiated equation may then fail to determine a finite value of \(\frac{dy}{dx}\); additional analysis is needed rather than dividing by zero.

This condition matters even when all the differentiation is correct. A derivative formula obtained by division is valid only at points where its denominator is nonzero, and the point must lie on the original relation.