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Computing Partial Derivatives · Tutorial 311 of 999

Zero Partials and Coordinate-Independent Terms

Learn to identify coordinate-independent terms, compute their zero partial derivatives, and distinguish genuine independence from dependence hidden inside an expression.

Beginner 9 min read

What You'll Learn

  • Define independence with respect to one coordinate while holding the others fixed.
  • Use the partial-derivative difference quotient to justify a zero partial.
  • Identify coordinate-independent terms inside larger expressions.
  • Recognize hidden dependence and cancellations in algebraic expressions.
  • Understand when zero partials imply independence along a coordinate interval.
  • Avoid treating a term as constant when it actually depends on the chosen coordinate.

Why Coordinate Independence Matters

A partial derivative measures change with respect to one coordinate while the other coordinates are held fixed. This makes it important to distinguish which parts of an expression can change as that coordinate varies and which parts cannot.

A term that is independent of the chosen coordinate contributes zero to the corresponding partial derivative. Recognizing such terms can simplify a calculation before any differentiation rules are applied. But the appearance of a symbol is not always decisive: a coordinate may be hidden inside a larger expression, or visible occurrences may cancel algebraically.

Key idea: Independence is always relative to a chosen coordinate. A function can depend on \(y\) and \(z\) while being independent of \(x\); its partial derivative with respect to \(x\) is then zero.

What Independence from a Coordinate Means

Let \(f\) be a function of coordinates \(x,y,\ldots\), defined on a domain \(D\). To say that \(f\) is independent of \(x\) means that changing \(x\), while leaving all the other coordinates unchanged, does not change the value of \(f\), whenever both resulting points lie in \(D\).

Definition: The function \(f\) is independent of \(x\) on \(D\) if, whenever two points in \(D\) agree in every coordinate except possibly \(x\), their \(f\)-values are equal. A term is independent of \(x\) when, viewed as a function, it has this property.

For example, \(y^2+3z\) is independent of \(x\). Its value does not change when \(x\) changes and \(y,z\) stay fixed. In contrast, \((x+y)^2\) depends on \(x\), even though \(x\) appears inside a grouped expression.

The domain matters. If a formula includes a denominator \(y-2\), the function is defined only where \(y\ne2\). Independence statements and partial derivatives apply where the function is defined, and difference quotients must use sufficiently small changes that remain in the domain.

Why an Independent Term Has Zero Partial

The partial derivative with respect to \(x\) is formed by changing \(x\) by a small increment \(h\) and keeping the other coordinates fixed. If the function is independent of \(x\), this change leaves its value exactly the same. The difference quotient is therefore zero for every allowed nonzero \(h\), so its limit is zero.

$$ f_x(x,y,\ldots) = \lim_{h\to 0} \frac{f(x+h,y,\ldots)-f(x,y,\ldots)}{h} = \lim_{h\to 0}\frac{0}{h} = 0. $$

This argument applies at an interior point where the partial derivative is defined. In particular, independence on a neighborhood of the point guarantees that sufficiently small changes in \(x\) are available and leave the function unchanged.

Result: If \(f\) is independent of \(x\) near a point and \(f_x\) is defined there, then \(f_x=0\) at that point. The other coordinates are held fixed throughout the difference quotient.

Worked Example: A Function with No \(x\)-Dependence

Worked Example: A Function with No \(x\)-Dependence

Let \(f(x,y)=5y^3-4y+9\), defined for all real \(x,y\). Find \(f_x\).

The expression contains no \(x\). If \(y\) is held fixed, changing \(x\) does not alter any part of \(f\). Using the difference quotient directly,

$$ \begin{aligned} f_x(x,y) &=\lim_{h\to0}\frac{f(x+h,y)-f(x,y)}{h}\\ &=\lim_{h\to0}\frac{(5y^3-4y+9)-(5y^3-4y+9)}{h}\\ &=\lim_{h\to0}\frac{0}{h}\\ &=0. \end{aligned} $$

Check: For every fixed \(y\), both function values in the numerator are the same expression, \(5y^3-4y+9\). Their difference is exactly zero, confirming \(f_x=0\).

Finding Independent Terms Inside an Expression

Often only some terms are independent of the selected coordinate. Earlier material on linearity of partial differentiation allows a sum to be differentiated term by term. An independent term contributes zero, while terms that do depend on the coordinate must still be differentiated using the appropriate rules.

For instance, when differentiating with respect to \(x\), \(y^2\) is independent of \(x\), but \(\sin(x+y)\) is not: changing \(x\) changes its input. The fact that \(y\) is held fixed does not make the whole expression \(x+y\) fixed.

Worked Example: Separating an Independent Term from a Dependent One

Let \(g(x,y)=\sin(x+y)+y^2\). Find \(g_x\).

With respect to \(x\), the term \(y^2\) is constant. The term \(\sin(x+y)\) depends on \(x\) through its input. By linearity and the chain rule,

$$ \begin{aligned} g_x(x,y) &=\frac{\partial}{\partial x}\sin(x+y) +\frac{\partial}{\partial x}y^2\\ &=\cos(x+y)\frac{\partial}{\partial x}(x+y)+0\\ &=\cos(x+y)\cdot 1\\ &=\cos(x+y). \end{aligned} $$

Check: In the difference quotient for the \(y^2\) term, the numerator is \(y^2-y^2=0\). For the sine term, changing \(x\) changes \(x+y\) by \(h\), so its quotient is \(\bigl(\sin(x+y+h)-\sin(x+y)\bigr)/h\), whose limit is \(\cos(x+y)\). These contributions give the stated result.

Visible Occurrences and Hidden Dependence

A coordinate can affect a term even when it does not appear as a separate factor. For example, \(e^{xz}\) depends on \(x\), because changing \(x\) changes the exponent whenever \(z\ne0\). On the other hand, \(e^z\) is independent of \(x\). The test is whether the term's value changes as the chosen coordinate varies, not whether the expression looks complicated.

Worked Example: Checking Dependence in a Larger Expression

For \(y\ne2\), let

$$ F(x,y,z)=\frac{x^2+1}{y-2}+e^{xz}+\cos y. $$

Find \(F_y\). The exponential term \(e^{xz}\) is independent of \(y\), so its partial derivative with respect to \(y\) is zero. The other two terms depend on \(y\). Differentiating them gives

$$ \begin{aligned} F_y(x,y,z) &=(x^2+1)\frac{\partial}{\partial y}(y-2)^{-1} +\frac{\partial}{\partial y}e^{xz} +\frac{\partial}{\partial y}\cos y\\ &=(x^2+1)\bigl(-(y-2)^{-2}\bigr)+0-\sin y\\ &=-\frac{x^2+1}{(y-2)^2}-\sin y. \end{aligned} $$

Check: In the difference quotient with \(x,z\) fixed, the exponential contribution is \(\bigl(e^{xz}-e^{xz}\bigr)/h=0\). For the first term, writing \(A=x^2+1\), its quotient is

$$ \frac{1}{h}\left(\frac{A}{y+h-2}-\frac{A}{y-2}\right) = -\frac{A}{(y+h-2)(y-2)} \longrightarrow -\frac{A}{(y-2)^2}. $$

The cosine contribution tends to \(-\sin y\). Together these limits confirm the formula. The restriction \(y\ne2\) is necessary because the original function is not defined at \(y=2\).

Algebra Can Reveal Independence

Sometimes a formula contains the chosen coordinate in several places, but those occurrences cancel. In that situation, inspecting the unsimplified expression can be misleading. Simplifying first can show whether the function truly depends on that coordinate.

Worked Example: Cancellation of Apparent \(y\)-Dependence

Let \(H(x,y,z)=(x+y)^2-2xy-y^2+3z^2\). Find \(H_y\).

Expanding the square gives

$$ \begin{aligned} H(x,y,z) &=x^2+2xy+y^2-2xy-y^2+3z^2\\ &=x^2+3z^2. \end{aligned} $$

The simplified expression is independent of \(y\). Therefore,

$$ H_y(x,y,z)=0. $$

Check: For fixed \(x,z\), the simplified value \(x^2+3z^2\) is unchanged when \(y\) changes. Thus the numerator in the difference quotient is \((x^2+3z^2)-(x^2+3z^2)=0\), and the quotient and its limit are zero. Differentiating the unsimplified terms also confirms the cancellation: \(2(x+y)-2x-2y=0\).

When Does a Zero Partial Imply Independence?

Independence implies a zero partial wherever that partial is defined. The converse needs a condition: a partial derivative equal to zero at just one point does not show that the function is independent of the coordinate. For example, \(q(x)=x^2\) has \(q'(0)=0\), but its values are not constant as \(x\) varies.

There is a useful converse along an interval. Fix all coordinates except \(x\), and suppose the allowed \(x\)-values form an interval \(I\). If the resulting single-variable function is differentiable throughout \(I\) and its derivative is zero everywhere there, then it is constant on \(I\). This follows from the Mean Value Theorem.

Theorem: Suppose the \(x\)-slice of a function is defined on an interval \(I\), is continuous on every closed subinterval of \(I\), and is differentiable in the interior of every such subinterval. If \(f_x=0\) throughout \(I\), then \(f\) is constant as \(x\) varies along that slice.

Proof. Choose any \(a,b\in I\) with \(a<b\), while keeping the other coordinates fixed. The Mean Value Theorem gives a point \(c\in(a,b)\) such that

$$ f(b,\ldots)-f(a,\ldots)=f_x(c,\ldots)(b-a). $$

By assumption, \(f_x(c,\ldots)=0\). Hence \(f(b,\ldots)-f(a,\ldots)=0\), so \(f(b,\ldots)=f(a,\ldots)\). Since the choice of \(a,b\) was arbitrary, the function is constant along the interval and is independent of \(x\) on that slice. This conclusion applies separately to each coordinate slice; on a domain with disconnected slices, it does not by itself establish equal values across different components.

A Common Mistake: Holding Fixed the Wrong Thing

A frequent error is to see \(y\) inside an expression and conclude that the entire expression is constant when differentiating with respect to \(x\). Only \(y\) and the other non-\(x\) coordinates are held fixed. Any occurrence of \(x\), including one inside a function's input, can still change.

Incorrect: \(\frac{\partial}{\partial x}\sin(x+y)=0\), on the grounds that \(y\) is held fixed.

Correction: Holding \(y\) fixed does not hold \(x+y\) fixed. Its \(x\)-derivative is \(1\), so \(\frac{\partial}{\partial x}\sin(x+y)=\cos(x+y)\). By contrast, \(\frac{\partial}{\partial x}\sin y=0\), because \(\sin y\) has no dependence on \(x\).

A reliable check is to imagine changing only the chosen coordinate by a small amount. If the term's value can change, it is not independent of that coordinate. If algebraic simplification shows the value cannot change, its partial derivative is zero wherever defined.

Why This Skill Is Useful

Recognizing coordinate-independent terms reduces unnecessary work and helps prevent accidental differentiation of the wrong variable. It also makes larger calculations easier to organize: identify the terms that change, set the independent terms' contributions to zero, and then apply the relevant differentiation rules to the remaining terms.

The essential distinction is between a term that is genuinely unchanged when one coordinate varies and a term that merely looks as though it might be. Hidden dependence, domain restrictions, and algebraic cancellation can all affect that judgment. The difference quotient gives a direct test: if the function's value does not change, the quotient is zero; if it does change, the term must be analyzed rather than discarded.

Check Your Understanding

For each question, identify the coordinate being varied and keep all other coordinates fixed.

  1. For \(f(x,y)=x^2+4y+1\), which partial derivative is zero, and why?
  2. Find the contribution of \(y^3\) to the partial derivative with respect to \(x\) of \(x\sin y+y^3\).
  3. Does \(e^{xz}\) depend on \(x\)? Does it depend on \(y\)? Explain your answers.
  4. Simplify \((x+y)^2-x^2-2xy\), then determine its partial derivative with respect to \(y\).
  5. Why does \(q'(0)=0\) alone not prove that \(q\) is independent of its variable? Give an example.

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