Difficulty: Intermediate–Advanced | Prerequisites: First-order partial derivatives, implicit differentiation from Calculus I.
Sometimes a relationship between x, y and z is given as an equation like F(x, y, z) = 0 rather than as z = f(x, y). You cannot simply isolate z and differentiate. Instead, you differentiate both sides of the equation with respect to your chosen variable, treating z as an implicit function of x and y, and solve for the partial derivative. This is exactly the technique you learnt for dy/dx in Calculus I, extended to multiple variables.
When you cannot isolate z in an equation involving x, y and z, you use implicit differentiation to find partial z / partial x and partial z / partial y. Differentiate both sides with respect to your chosen variable, attach a partial z / partial x (or partial z / partial y) to every term involving z via the chain rule, then solve algebraically for the partial derivative.
Implicitly defined function
A function z = f(x, y) that is defined by an equation F(x, y, z) = 0 (or some constant) rather than by an explicit formula z = something. You know z depends on x and y, but you cannot (or choose not to) solve for z directly.
Think of it as: z is hiding inside the equation, and you have to tease out how it changes without ever pulling it out.
Implicit partial differentiation
The technique of differentiating both sides of an implicit equation with respect to one variable (say x), applying the chain rule to every occurrence of z (since z depends on x), and then solving for partial z / partial x.
In simple terms, this means: you treat z as a function of x and y even though you cannot write it down explicitly, and use the chain rule to account for that hidden dependence.
Start with the equation F(x, y, z) = constant. Understand that z is the "output" (a function of x and y), even though you cannot solve for it.
Differentiate every term of both sides with respect to the variable you want (x for partial z / partial x, y for partial z / partial y).
Wherever z appears, apply the chain rule: the derivative of a term involving z picks up a factor of partial z / partial x (or partial z / partial y).
Collect all terms containing partial z / partial x on one side, move everything else to the other side.
Factor out partial z / partial x and solve.
Given: x^3 + y^3 + z^3 + 6xyz = 1. Find partial z / partial x.
Differentiate both sides with respect to x, treating y as constant and z as z(x, y):
d/dx of x^3 = 3x^2
d/dx of y^3 = 0 (y is constant with respect to x)
d/dx of z^3 = 3z^2 (partial z / partial x), by the chain rule
d/dx of 6xyz = 6yz + 6xy (partial z / partial x), by the product rule (z depends on x)
Putting it together:
3x^2 + 0 + 3z^2 (partial z / partial x) + 6yz + 6xy (partial z / partial x) = 0
Collect terms with partial z / partial x:
3x^2 + 6yz = -(3z^2 + 6xy)(partial z / partial x)
Solve:
partial z / partial x = (3x^2 + 6yz) / (-3z^2 - 6xy)
This simplifies to -(x^2 + 2yz) / (z^2 + 2xy).
Given: x - z = arctan(yz). Find partial z / partial x and partial z / partial y.
For partial z / partial x, differentiate with respect to x:
Left side: 1 - partial z / partial x
Right side: the derivative of arctan(yz) with respect to x = [1 / (1 + (yz)^2)] times y (partial z / partial x), since y is constant and z depends on x
So: 1 - partial z / partial x = y (partial z / partial x) / (1 + (yz)^2)
Solving for partial z / partial x:
partial z / partial x = 1 / [y / (1 + (yz)^2) + 1]
For partial z / partial y, differentiate with respect to y:
Left side: 0 - partial z / partial y = -partial z / partial y
Right side: [1 / (1 + (yz)^2)] times (z + y (partial z / partial y)), using the product rule on yz
Solving for partial z / partial y:
partial z / partial y = [-z / (1 + (yz)^2)] / [y / (1 + (yz)^2) + 1]
For an equation F(x, y, z) = c where z = z(x, y), the general pattern is:
\frac{\partial z}{\partial x} = -\frac{F_x}{F_z}\frac{\partial z}{\partial y} = -\frac{F_y}{F_z}where F_x, F_y, F_z are the partial derivatives of F with respect to x, y, z respectively, and F_z is nonzero. This shortcut formula comes from the implicit function theorem. You can use it directly or derive the result each time by the step-by-step method above, whichever feels more reliable during an exam.
Students forget to apply the chain rule to terms containing z. Every time you differentiate a term with z in it, a factor of partial z / partial x (or partial z / partial y) must appear. Missing this is the single most common error.
When differentiating 6xyz with respect to x, students sometimes forget that z also depends on x. The product rule applies: d/dx of (6xyz) = 6yz + 6xy (partial z / partial x). Both terms are needed.
Students occasionally confuse which variable is constant. When finding partial z / partial x, y is constant, not z. And z is not constant either: z is the unknown function you are solving for.
Some students try to use the shortcut formula -F_x / F_z without understanding where it comes from, and then apply it to equations where the setup does not match F(x, y, z) = c. If you are unsure, do the step-by-step method instead.
Implicit partial differentiation is tested directly: expect an equation involving x, y and z with instructions to find partial z / partial x and partial z / partial y.
The algebraic manipulation (collecting terms, factoring out the partial derivative) is where most marks are lost. Show every step clearly.
This technique is a prerequisite for the gradient of a level surface and for understanding tangent planes to implicitly defined surfaces.
Both the step-by-step method and the -F_x / F_z shortcut are valid on exams. If your professor taught one explicitly, use that one.
True or false: When finding partial z / partial x from x^3 + y^3 + z^3 = 1, you treat y as a constant and z as a function of x.
True.
Fill in the blank: When you differentiate z^3 with respect to x (where z depends on x), you get ______.
3z^2 (partial z / partial x).
True or false: The shortcut formula for implicit partial derivatives is partial z / partial x = F_x / F_z.
False. It is -F_x / F_z (note the negative sign).
Fill in the blank: In the product rule applied to d/dx of (6xyz), treating y as constant and z as z(x, y), the result is ______.
6yz + 6xy (partial z / partial x).
Q: Given x^3 + y^3 + z^3 + 6xyz = 1, find partial z / partial x.
A: Differentiate with respect to x: 3x^2 + 3z^2 (partial z / partial x) + 6yz + 6xy (partial z / partial x) = 0. Rearranging: partial z / partial x = -(3x^2 + 6yz) / (3z^2 + 6xy), which simplifies to -(x^2 + 2yz) / (z^2 + 2xy).
Q: Given x - z = arctan(yz), find partial z / partial x.
A: Differentiate with respect to x: 1 - partial z / partial x = [y (partial z / partial x)] / (1 + (yz)^2). Solving: partial z / partial x = 1 / [y / (1 + (yz)^2) + 1], which can also be written as (1 + (yz)^2) / (1 + (yz)^2 + y).
Q: Use the shortcut formula to find partial z / partial x for x^3 + y^3 + z^3 + 6xyz = 1.
A: Let F(x, y, z) = x^3 + y^3 + z^3 + 6xyz - 1. Then F_x = 3x^2 + 6yz, F_z = 3z^2 + 6xy. The shortcut gives partial z / partial x = -F_x / F_z = -(3x^2 + 6yz) / (3z^2 + 6xy). Same answer as the step-by-step method.
Implicit partial differentiation is the gateway to understanding level surfaces and the gradient vector for implicitly defined functions. The gradient of F at a point on the surface F(x, y, z) = c is normal to the surface, which leads directly to equations for tangent planes and normal lines.
In later courses, the implicit function theorem formalises the conditions under which you can treat z as a function of x and y, and the technique extends to systems of equations with multiple implicitly defined functions.
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