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Implicit Differentiation for Surfaces

Use the chain rule and the implicit function theorem to find and verify the partial derivatives of a surface locally written as \(z=z(x,y)\).

Intermediate 9 min read

What You'll Learn

  • Identify when an equation \(F(x,y,z)=0\) defines \(z\) locally as a function of \(x\) and \(y\).
  • Derive the formulas for \(z_x\) and \(z_y\) using the multivariable chain rule.
  • Evaluate the formulas at a point on the surface.
  • Check that a specified point satisfies the surface equation and the derivative identities.
  • Recognize why the formulas cannot be used when \(F_z=0\).

From an Equation to a Local Surface Graph

A surface may be described by an equation relating three variables, such as \(F(x,y,z)=0\). Sometimes the equation can be solved explicitly for \(z\), but often that is inconvenient or impossible in a useful form. Implicit differentiation provides another route: when the surface can locally be viewed as a graph \(z=z(x,y)\), the chain rule gives its partial derivatives directly from \(F\).

The word locally matters. A surface might fail to be the graph of a single-valued function over all \(x\)- and \(y\)-values, yet still have that form near a particular point. The key condition for the standard local result is that the derivative of \(F\) with respect to \(z\) is nonzero at that point.

Definition: The level surface defined by \(F(x,y,z)=0\) is the set of points \((x,y,z)\) satisfying that equation. A local graph representation near \((a,b,c)\) is a differentiable function \(z=z(x,y)\), defined for \((x,y)\) near \((a,b)\), such that \(z(a,b)=c\) and \(F(x,y,z(x,y))=0\).

The Implicit Partial-Derivative Formulas

Suppose \(F\) has continuous first partial derivatives near \((a,b,c)\), the point lies on the surface, and \(F_z(a,b,c)\ne 0\). The implicit function theorem guarantees that, near this point, the surface can be represented uniquely as a differentiable function \(z=z(x,y)\). The subscripts on \(F\) denote partial derivatives taken with the other two coordinates held fixed.

Theorem: Under those conditions, at any nearby point on the local graph where \(F_z\ne 0\),
$$ z_x=-\frac{F_x}{F_z}, \qquad z_y=-\frac{F_y}{F_z}. $$

All three partial derivatives on the right are evaluated at the same point \((x,y,z(x,y))\). The formula for \(z_x\) describes how \(z\) changes as \(x\) changes while \(y\) is held fixed. The formula for \(z_y\) describes how \(z\) changes as \(y\) changes while \(x\) is held fixed.

Why the Formulas Work

Once the surface is written locally as a graph, define the composite function \(G(x,y)=F(x,y,z(x,y))\). Every point on the graph lies on the surface, so \(G(x,y)=0\) throughout its domain. Differentiating this identity with respect to \(x\), while holding \(y\) fixed, applies the chain rule to each way \(x\) affects \(F\): directly through its first argument and indirectly through \(z(x,y)\).

$$ \frac{\partial}{\partial x}F(x,y,z(x,y)) = F_x+F_z z_x = 0. $$

Since \(F_z\ne 0\), rearranging gives \(z_x=-F_x/F_z\). Differentiating the same identity with respect to \(y\), while holding \(x\) fixed, gives

$$ \frac{\partial}{\partial y}F(x,y,z(x,y)) = F_y+F_z z_y = 0, \qquad z_y=-\frac{F_y}{F_z}. $$

This also explains why \(F_z\) appears in the denominator: changes in \(z\) must compensate for the direct change in \(F\) caused by varying \(x\) or \(y\). A nonzero \(F_z\) makes that compensation solvable by division.

Worked Example: A Point on a Sphere

Worked Example: A Point on a Sphere

Find \(z_x\) and \(z_y\) for the sphere \(x^2+y^2+z^2=14\) at \((1,2,3)\), treating \(z\) locally as a function of \(x\) and \(y\).

First define \(F(x,y,z)=x^2+y^2+z^2-14\). Check that the given point lies on the surface:

$$ F(1,2,3)=1^2+2^2+3^2-14=1+4+9-14=0. $$

The required partial derivatives of \(F\) are

$$ F_x=2x,\qquad F_y=2y,\qquad F_z=2z. $$

At \((1,2,3)\), these take the values \(F_x=2\), \(F_y=4\), and \(F_z=6\). Since \(F_z=6\ne 0\), the local graph condition holds, and

$$ z_x=-\frac{2}{6}=-\frac{1}{3}, \qquad z_y=-\frac{4}{6}=-\frac{2}{3}. $$

Check both results in the differentiated surface identities. For the \(x\)-partial, \(F_x+F_z z_x=2+6(-1/3)=0\). For the \(y\)-partial, \(F_y+F_z z_y=4+6(-2/3)=0\). Both are consistent with differentiating \(F=0\).

Worked Example: A Mixed Polynomial Equation

Worked Example: A Mixed Polynomial Equation

Let \(F(x,y,z)=xz+y^2+z^3-3\). Find \(z_x\) and \(z_y\) at \((1,1,1)\).

Verify the point first:

$$ F(1,1,1)=1\cdot1+1^2+1^3-3=1+1+1-3=0. $$

Differentiate \(F\) with respect to each coordinate, holding the other coordinates fixed:

$$ F_x=z,\qquad F_y=2y,\qquad F_z=x+3z^2. $$

At the specified point, \(F_x=1\), \(F_y=2\), and \(F_z=1+3(1)^2=4\). The denominator is nonzero, so

$$ z_x=-\frac{1}{4}, \qquad z_y=-\frac{2}{4}=-\frac{1}{2}. $$

Substitution checks the differentiated identities: \(F_x+F_z z_x=1+4(-1/4)=0\), and \(F_y+F_z z_y=2+4(-1/2)=0\). Thus both partial derivatives satisfy the equations obtained from \(F(x,y,z(x,y))=0\).

Worked Example: An Exponential Surface

Worked Example: An Exponential Surface

Consider \(F(x,y,z)=e^z+xz+y^2-2\). Find \(z_x\) and \(z_y\) at \((0,1,0)\).

The point is on the surface because

$$ F(0,1,0)=e^0+(0)(0)+1^2-2=1+0+1-2=0. $$

The partial derivatives are

$$ F_x=z,\qquad F_y=2y,\qquad F_z=e^z+x. $$

At \((0,1,0)\), this gives \(F_x=0\), \(F_y=2\), and \(F_z=e^0+0=1\). Since \(F_z\ne 0\), the formulas apply:

$$ z_x=-\frac{0}{1}=0, \qquad z_y=-\frac{2}{1}=-2. $$

For the check, \(F_x+F_z z_x=0+1(0)=0\), while \(F_y+F_z z_y=2+1(-2)=0\). The values therefore satisfy both differentiated identities at the specified point.

A Reliable Calculation Process

The same sequence of checks keeps implicit differentiation organized for many different surface equations.

1
Write the equation as \(F=0\).
Move all terms to one side and define \(F(x,y,z)\) explicitly.
2
Check the specified point.
Substitute its three coordinates into \(F\); the result must be zero.
3
Compute \(F_x\), \(F_y\), and \(F_z\).
These derivatives treat \(x,y,z\) as independent coordinates when they are computed.
4
Check the denominator.
Evaluate \(F_z\) at the point. The stated local graph theorem applies if this value is nonzero.
5
Evaluate both quotients and verify.
Find \(-F_x/F_z\) and \(-F_y/F_z\), then substitute them into \(F_x+F_z z_x=0\) and \(F_y+F_z z_y=0\).

What the Condition \(F_z\ne 0\) Does—and Does Not—Say

The nonzero condition is what allows the implicit function theorem to guarantee a unique differentiable graph \(z=z(x,y)\) near the point and allows the formulas to be rearranged by division. It is important to evaluate \(F_z\) at the point in question, not merely to note that \(F_z\) is not identically zero elsewhere on the surface.

If \(F_z=0\) at a point, the displayed quotient formulas are unavailable there. That does not by itself prove that no local graph exists; it only means this condition does not guarantee one and this formula cannot be used to compute its partial derivatives. For instance, \(F(x,y,z)=z\) defines the plane \(z=0\), which is a graph, even though \(F_z=1\) actually nonzero; a more relevant zero derivative example is \(F(x,y,z)=z^2\), whose zero set is also \(z=0\) but has \(F_z=2z=0\) on that set. Thus the nonzero condition is sufficient for the theorem, not a necessary condition for every possible graph representation.

A Common Sign and Chain-Rule Mistake

A frequent error is to differentiate the surface equation with respect to \(x\) and forget that \(z\) changes when \(x\) changes. On the sphere \(F=x^2+y^2+z^2-14=0\), an incorrect calculation might write \(2x=0\), as if \(z\) were constant, and conclude that the equation imposes no meaningful derivative information. That calculation omits the dependence \(z=z(x,y)\).

The correct \(x\)-partial differentiates both the direct \(x\)-term and the \(z\)-term:

$$ 2x+2z z_x=0. $$

At \((1,2,3)\), this becomes \(2+6z_x=0\), so \(z_x=-1/3\). The analogous \(y\)-partial is \(2y+2z z_y=0\), giving \(4+6z_y=0\) and \(z_y=-2/3\). In each case, the additional factor \(z_x\) or \(z_y\) is the chain-rule contribution from the changing height of the surface.

Interpreting the Results

A partial derivative gives a local rate of change of height. For example, \(z_x=-1/3\) at the sphere point means that, near \((1,2)\), increasing \(x\) while keeping \(y\) fixed changes the corresponding surface height at an instantaneous rate of \(-1/3\) height units per \(x\)-unit. The value of \(z_y\) describes the corresponding change when \(y\) varies and \(x\) is fixed.

These are local statements: they describe the surface near the point, not a claim that the same rate holds everywhere. The formulas are useful precisely because they turn a possibly complicated surface equation into a calculation using its first partial derivatives, while making the necessary graph condition explicit.

Check Your Understanding

For each question, keep track of which coordinate is being held fixed and check whether the local graph condition applies.

  1. For \(F(x,y,z)=x^2+y^2+z^2-9\), verify that \((2,1,2)\) lies on the surface, then find \(z_x\) and \(z_y\) there.
  2. For a local graph satisfying \(F(x,y,z(x,y))=0\), what equation results from differentiating with respect to \(y\)?
  3. Why must \(F_z\) be nonzero to use the quotient formula for \(z_x\) at a specified point?
  4. At a point where \(F_x=3\) and \(F_z=-2\), what is \(z_x\), assuming the local graph conditions hold?
  5. Does \(F_z=0\) prove that the surface cannot be represented locally as \(z=z(x,y)\)? Explain.

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