Power Rules for Multivariable Monomials

One Variable Changes at a Time

A multivariable monomial can involve several coordinates, but a partial derivative changes only one coordinate at a time. When differentiating with respect to \(x\), for example, the factors involving \(y\) and \(z\) do not change. At a point where those factors are defined, they act as constants in the one-variable calculation.

This observation lets us use the ordinary power rule on one factor. The main details to watch are the exponent on the factor being differentiated and the points where the original monomial is defined. These details matter especially for zero and negative exponents.

Definition: For integer exponents \(m,n,p\), a three-variable monomial is an expression \(f(x,y,z)=x^m y^n z^p\). A factor with exponent zero is defined to be \(1\), including when its base is zero. A factor with a negative integer exponent is defined only when its base is nonzero.

Thus, if \(m<0\), the domain requires \(x\ne 0\); if \(n<0\), it requires \(y\ne 0\); and if \(p<0\), it requires \(z\ne 0\). Positive and zero exponents impose no such restrictions. The same conventions apply to monomials in two variables, such as \(x^m y^n\).

The Partial Power Rule

Recall that a partial derivative with respect to \(x\) differentiates the function as \(x\) varies while the other coordinates are held fixed. In the difference quotient, this means replacing \(x\) by \(x+h\), while keeping \(y\) and \(z\) unchanged.

Rule: For \(f(x,y,z)=x^m y^n z^p\), where \(m,n,p\) are integers, the partial derivative with respect to \(x\) is \[ \frac{\partial f}{\partial x}=m x^{m-1}y^n z^p. \] The formula applies at points in the domain of \(f\). If \(m=0\), the partial derivative is zero; interpret that case as differentiating a constant factor, rather than evaluating an expression such as \(0x^{-1}\) at \(x=0\). The corresponding rules for \(y\) and \(z\) are obtained by lowering the exponent of the variable being differentiated and leaving the other factors unchanged.

In particular, when the relevant formula is defined, the three partial derivatives are

\frac{\partial f}{\partial x}=m x^{m-1}y^n z^p,\qquad \frac{\partial f}{\partial y}=n x^m y^{n-1}z^p,\qquad \frac{\partial f}{\partial z}=p x^m y^n z^{p-1}.

“When the relevant formula is defined” is important. If \(m=0\), the expression \(m x^{m-1}\) may appear to involve \(x^{-1}\), even though the original factor \(x^0\) is simply the constant \(1\). The correct derivative is then zero, including at \(x=0\), if the other factors are defined there. If \(m>0\), the usual formula is defined at \(x=0\) as well: when \(m=1\), the factor \(x^{m-1}\) is \(x^0=1\); when \(m>1\), it is a positive power. If \(m<0\), the original function is defined only for \(x\ne0\), and so is its derivative formula.

Why the Rule Works

Fix values of \(y\) and \(z\) so the monomial is defined, and call their product \(C=y^n z^p\). As \(x\) varies, the function is the one-variable function \(C x^m\). Its difference quotient is \(C\) times the difference quotient for \(x^m\). For positive integer exponents, factoring the difference of powers shows that quotient tends to \(m x^{m-1}\). For negative integer exponents, the same result follows by writing the power as a reciprocal and taking the limit.

Freeze the other coordinates.
Set \(C=y^n z^p\). For the partial derivative with respect to \(x\), \(C\) is constant.

Use the difference quotient.
For \(x\) in the domain, the quotient for \(Cx^m\) is \(C\big((x+h)^m-x^m\big)/h\).

Take the one-variable limit.
The power quotient tends to \(m x^{m-1}\), so the partial derivative is \(C m x^{m-1}\).

Here is the power-quotient calculation in detail. For a positive integer \(m\), the difference-of-powers identity gives

\frac{(x+h)^m-x^m}{h} =\sum_{j=0}^{m-1}(x+h)^{m-1-j}x^j \longrightarrow \sum_{j=0}^{m-1}x^{m-1} =m x^{m-1}.

For a negative integer \(m=-k\), where \(k\) is a positive integer, \(x\ne0\), and \(h\) is sufficiently small that \(x+h\ne0\). Then

\frac{(x+h)^{-k}-x^{-k}}{h} = -\frac{((x+h)^k-x^k)/h}{x^k(x+h)^k} \longrightarrow -\frac{kx^{k-1}}{x^{2k}} =-k x^{-k-1} =m x^{m-1}.

When \(m=0\), \(x^m=1\), so its difference quotient is \(0\). Multiplying these one-variable results by the fixed factor \(C\) proves the partial power rule. The same reasoning applies when the changing coordinate is \(y\) or \(z\).

Worked Examples

Worked Example: Three Positive Exponents

Let \(f(x,y,z)=x^4y^2z^3\). All exponents are positive, so the domain is all of \(\mathbb{R}^3\).

For the \(x\)-partial, lower the exponent of \(x\) by one and multiply by its original exponent. Keep the other factors unchanged:

\frac{\partial f}{\partial x} =4x^{4-1}y^2z^3 =4x^3y^2z^3.

For the \(y\)-partial, lower the exponent of \(y\):

\frac{\partial f}{\partial y} =2x^4y^{2-1}z^3 =2x^4yz^3.

For the \(z\)-partial:

\frac{\partial f}{\partial z} =3x^4y^2z^{3-1} =3x^4y^2z^2.

Check: Fix \(y,z\). The resulting one-variable function is \(y^2z^3x^4\), whose derivative with respect to \(x\) is \(4y^2z^3x^3\), matching the first result. Fixing the other two coordinates in turn gives \(2x^4z^3y\) and \(3x^4y^2z^2\), matching the other results.

Worked Example: A Zero Exponent

Consider \(g(x,y,z)=x^3y^0z^2\). Since \(y^0=1\), this is the function \(x^3z^2\), defined for all real \(x,y,z\). Differentiate with respect to \(x\):

\frac{\partial g}{\partial x} =3x^{3-1}y^0z^2 =3x^2z^2.

With respect to \(y\), the function does not change at all, because its value is independent of \(y\). Therefore \(\partial g/\partial y=0\). In the formal power-rule expression, the coefficient is the exponent \(0\); it is the constant factor \(y^0=1\) that is being differentiated.

\frac{\partial g}{\partial y}=0.

With respect to \(z\), lower the exponent of \(z\):

\frac{\partial g}{\partial z} =2x^3y^0z^{2-1} =2x^3z.

Check: The \(y\)-slice is constant, \(x^3z^2\), so its derivative is \(0\). The \(x\)-slice is a constant multiple \(z^2x^3\), with derivative \(3z^2x^2\); the \(z\)-slice is \(x^3z^2\), with derivative \(2x^3z\). These agree with all three results, including at \(y=0\).

Worked Example: A Negative Exponent

Let \(q(x,y,z)=x^{-2}y^3z^0\). The factor \(x^{-2}=1/x^2\) requires \(x\ne0\), while \(z^0=1\) imposes no restriction. The domain is therefore the set of points with \(x\ne0\). For the \(x\)-partial, lower the exponent of \(x\):

\frac{\partial q}{\partial x} =(-2)x^{-2-1}y^3z^0 =-2x^{-3}y^3 =-\frac{2y^3}{x^3}.

For the \(y\)-partial:

\frac{\partial q}{\partial y} =3x^{-2}y^{3-1}z^0 =3x^{-2}y^2 =\frac{3y^2}{x^2}.

For the \(z\)-partial, the function is independent of \(z\), so \(\partial q/\partial z=0\).

\frac{\partial q}{\partial z}=0.

Check: With \(y,z\) fixed, the \(x\)-slice is \(y^3/x^2\), whose derivative for \(x\ne0\) is \(-2y^3/x^3\). With \(x,z\) fixed, the \(y\)-slice is \(y^3/x^2\), with derivative \(3y^2/x^2\). The \(z\)-slice is constant. Each slice confirms the stated partial derivative on the domain \(x\ne0\).

A Common Mistake: Changing More Than One Factor

A partial derivative changes one variable, not every variable in the expression. For example, if \(r(x,y)=x^2y^3\), an incorrect calculation of the \(x\)-partial might lower both exponents and produce \(6xy^2\). That calculation has changed the \(y\)-factor as though \(y\) were also varying. It is not the \(x\)-partial.

\text{Incorrect: }\frac{\partial}{\partial x}(x^2y^3)=6xy^2 \qquad\qquad \text{Correct: }\frac{\partial}{\partial x}(x^2y^3)=2xy^3.

To see the correction directly, fix \(y\). The function of \(x\) is \(y^3x^2\), so its derivative is \(y^3(2x)=2xy^3\). The \(y\)-factor remains unchanged. If differentiating with respect to \(y\), it is instead \(x^2\) that remains fixed, giving \(3x^2y^2\).

Using the Rule Reliably

For each partial derivative of a monomial, identify the variable being differentiated, multiply by that variable’s exponent, lower only that exponent by one, and preserve the other factors. Then check the domain: any variable with a negative exponent must be nonzero. If the exponent of the differentiated variable is zero, its factor is constant and the corresponding partial derivative is zero.

These rules concern a single monomial. They do not require differentiating a sum or using a product rule. Keeping the task to one factor at a time makes the calculation simple, while checking the exponent and domain prevents most errors.

Check Your Understanding

For each question, state any domain restriction before differentiating.

Find all three partial derivatives of \(f(x,y,z)=x^2y^4z\).
For \(g(x,y,z)=x^0y^2z^{-1}\), find the partial derivative with respect to \(x\). At which points is \(g\) defined?
Find \(\partial h/\partial y\) for \(h(x,y,z)=x^{-3}y^5z^2\), and state where the result is defined.
Explain why differentiating \(x^3y^2\) with respect to \(x\) does not lower the exponent of \(y\).
For \(k(x,y,z)=x^2y^0z^0\), determine \(\partial k/\partial z\), including whether the answer is defined when \(z=0\).