Partial Derivatives — Compute partial derivatives and higher-order partials - Multivariable Calculus

SECTION 1

Historical Context & Motivation

In single-variable calculus, you learned to find the derivative of a function like f(x) = x². But what happens when a quantity depends on more than one variable? For example, the temperature on a metal plate depends on both the x-position and the y-position. The volume of a cylinder depends on both its radius and its height. Scientists and mathematicians needed a way to study how a function changes when you adjust just one of those variables while holding the others fixed.

The concept of partial derivatives arose over centuries as mathematicians encountered problems in physics, engineering, and geometry that involved multiple changing quantities. Today, partial derivatives are fundamental tools used in fields from machine learning to weather forecasting.

1734 • Euler's Early Work

Leonhard Euler began using notation for functions of several variables, laying the groundwork for distinguishing between derivatives taken with respect to different variables.

1788 • Lagrange's Analytical Mechanics

Joseph-Louis Lagrange published his Mécanique analytique, which relied heavily on partial derivatives to describe the motion of complex mechanical systems with many moving parts.

1800s • The ∂ Notation Emerges

Adrien-Marie Legendre introduced the curly-d symbol ∂ (distinct from the ordinary d) to make it clear when a derivative is taken with respect to one variable while others are held constant.

1822 • Fourier's Heat Equation

Joseph Fourier published his theory of heat conduction, which used partial derivatives to model how temperature changes over both space and time — one of the most famous partial differential equations in history.

Modern • Machine Learning & Beyond

Today, partial derivatives power the gradient descent algorithm used to train neural networks, optimize engineering designs, and model climate systems — making them one of the most applied tools in modern mathematics.

The central question this lesson addresses is: How do we measure the rate of change of a function with respect to one variable while keeping all other variables constant? Answering this question opens the door to understanding surfaces, optimization, and the behavior of real-world systems.

SECTION 2

Core Principles & Definitions

A partial derivative measures how a multivariable function changes as you vary one input while treating all other inputs as constants. If you already know how to take an ordinary derivative, you already know most of what you need — the twist is simply deciding which variable to differentiate with respect to and freezing the rest.

One Variable at a Time

When computing ∂f/∂x, treat every variable except x as a constant. You apply the same derivative rules (power rule, product rule, chain rule) you already know from single-variable calculus.

The Curly-d Symbol ∂

The symbol ∂ ("partial") replaces the ordinary d to signal that other variables are being held constant. You read ∂f/∂x as "the partial derivative of f with respect to x."

Multiple Partials per Function

A function f(x, y) has two first-order partials: ∂f/∂x and ∂f/∂y. A function of three variables f(x, y, z) has three. Each one captures a different direction of change.

Higher-Order Partials

You can differentiate again to get second-order (and beyond) partials like ∂²f/∂x², ∂²f/∂y², or the mixed partial ∂²f/∂x∂y. These describe how the rate of change itself is changing.

Clairaut's Theorem

For most well-behaved functions, the mixed partial derivatives are equal: ∂²f/∂x∂y = ∂²f/∂y∂x. This powerful result means the order of differentiation usually doesn't matter.

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KEY TAKEAWAY

Think of a partial derivative like adjusting one dial on a mixing board while leaving all the other dials untouched. If a song's sound depends on bass, treble, and volume, a partial derivative with respect to bass tells you how the sound changes when only the bass knob moves — treble and volume stay fixed. Each dial gives you a different partial derivative, and together they describe the full picture of how the output responds to its inputs.

SECTION 3

Visual Explanation

The best way to understand partial derivatives is to visualize a surface in three dimensions. Imagine a hilly landscape described by a function z = f(x, y). At any point on that surface, you can walk in the x-direction (east-west) or the y-direction (north-south). The partial derivative ∂f/∂x tells you the slope of the hill if you walk purely east-west, and ∂f/∂y tells you the slope if you walk purely north-south.

The surface z = f(x, y) is shown as a semi-transparent shape. The pink dashed line represents the curve obtained by holding y constant and varying x; its slope at point P is ∂f/∂x. The amber dashed line represents the curve at fixed x; its slope at P is ∂f/∂y.

Notice that each slice through the surface produces a regular two-dimensional curve. Once you have that curve, you just take the ordinary derivative of it — that's the partial derivative. The key insight is that each partial derivative reduces a multivariable problem to a single-variable problem by freezing everything else.

SECTION 4

Mathematical Framework

Let's build the notation and rules step by step. If you can differentiate single-variable functions using the power rule, product rule, and chain rule, you can compute partial derivatives — the only new idea is treating other variables as constants.

FIRST-ORDER PARTIAL WITH RESPECT TO x

∂f/∂x = lim(h→0) [f(x + h, y) − f(x, y)] / h

This is the limit definition. Notice it looks exactly like the ordinary derivative limit, except y stays fixed. In practice, you just differentiate f with respect to x while treating y as a constant.

FIRST-ORDER PARTIAL WITH RESPECT TO y

∂f/∂y = lim(h→0) [f(x, y + h) − f(x, y)] / h

Same idea, but now x is held constant and y varies. Together, ∂f/∂x and ∂f/∂y describe the slope of the surface in two perpendicular directions.

Alternative Notation

You will encounter several ways to write partial derivatives. They all mean the same thing. For a function f(x, y):

Notation	Read as	Meaning
`∂f/∂x`	"partial f partial x"	Derivative of f treating y as constant
`fₓ(x, y)`	"f sub x"	Subscript notation for the same partial
`∂²f/∂x²`	"second partial of f with respect to x"	Differentiate with respect to x twice
`∂²f/∂y∂x`	"mixed partial: first x, then y"	Differentiate first with respect to x, then with respect to y

Common notation for partial derivatives

SECOND-ORDER PARTIALS

∂²f/∂x² = fₓₓ , ∂²f/∂y² = f_yy , ∂²f/∂y∂x = fₓᵧ

Second-order partials come in two flavors: unmixed (differentiate twice with respect to the same variable) and mixed (differentiate with respect to two different variables).

CLAIRAUT'S THEOREM

If fₓᵧ and f_yx are continuous, then ∂²f/∂y∂x = ∂²f/∂x∂y

This means the order in which you take mixed partial derivatives does not matter, as long as the partials are continuous — which is the case for nearly every function you'll encounter in this course.

SECTION 5

Higher-Order Partials in Detail

Once you know how to compute first-order partial derivatives, finding higher-order partial derivatives is simply a matter of repeating the process. You take the partial derivative of a partial derivative. For a function f(x, y), there are four second-order partials: fₓₓ, f_yy, fₓᵧ, and f_yx. Thanks to Clairaut's theorem, fₓᵧ and f_yx are usually equal, so in practice you compute three distinct second-order partials.

This tree diagram shows how a function f(x, y) branches into its first-order partials (fₓ and fᵧ), which then branch into four second-order partials. The green dashed box highlights Clairaut's theorem: the two mixed partials are equal.

Let's see a concrete example. Consider f(x, y) = x³y² + 2xy. Here are all first- and second-order partials:

fₓ = 3x²y² + 2y (treat y² and y as constants, differentiate x³ and x)
fᵧ = 2x³y + 2x (treat x³ and x as constants, differentiate y² and y)
fₓₓ = 6xy² (differentiate fₓ with respect to x again)
fᵧᵧ = 2x³ (differentiate fᵧ with respect to y again)
fₓᵧ = 6x²y + 2 (differentiate fₓ with respect to y)
fᵧₓ = 6x²y + 2 (differentiate fᵧ with respect to x — same as fₓᵧ ✓)

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Quick Check

Whenever you compute both mixed partials, verify that fₓᵧ = fᵧₓ. If they don't match, go back and check your algebra — for polynomial and trigonometric functions, they should always agree.

SECTION 6

Worked Example

Let's work through a complete example, finding all first-order and second-order partial derivatives for a function that mixes polynomial and trigonometric terms.

Find all first- and second-order partials of f(x, y) = x²sin(y) + 3y

1
Step 1 — Find ∂f/∂x

Treat y (and therefore sin(y)) as a constant. Differentiate x²sin(y) + 3y with respect to x. The term x²sin(y) differentiates to 2x·sin(y) using the power rule. The term 3y is a constant with respect to x, so its derivative is 0.

∂f/∂x = 2x·sin(y)
2
Step 2 — Find ∂f/∂y

Now treat x as a constant. Differentiate x²sin(y) + 3y with respect to y. The term x²sin(y) becomes x²cos(y) because the derivative of sin(y) is cos(y). The term 3y becomes 3.

∂f/∂y = x²cos(y) + 3
3
Step 3 — Find ∂²f/∂x² (fₓₓ)

Take the result from Step 1 (fₓ = 2x·sin(y)) and differentiate it with respect to x again. Since sin(y) is still a constant with respect to x, the derivative is 2·sin(y).

∂²f/∂x² = 2sin(y)
4
Step 4 — Find ∂²f/∂y² (fᵧᵧ)

Take the result from Step 2 (fᵧ = x²cos(y) + 3) and differentiate with respect to y again. The derivative of x²cos(y) is −x²sin(y) (since the derivative of cos(y) is −sin(y)). The derivative of the constant 3 is 0.

∂²f/∂y² = −x²sin(y)
5
Step 5 — Find the mixed partial ∂²f/∂y∂x (fₓᵧ)

Take fₓ = 2x·sin(y) and differentiate with respect to y. The derivative of sin(y) is cos(y), and 2x is treated as a constant, giving 2x·cos(y).

∂²f/∂y∂x = 2x·cos(y)
6
Step 6 — Verify Clairaut's Theorem

Take fᵧ = x²cos(y) + 3 and differentiate with respect to x. The derivative of x²cos(y) with respect to x is 2x·cos(y), and the constant 3 disappears. We get fᵧₓ = 2x·cos(y), which matches fₓᵧ. Clairaut's theorem is confirmed.

∂²f/∂x∂y = 2x·cos(y) = ∂²f/∂y∂x ✓

SECTION 7

Common Mistakes & How to Avoid Them

Partial derivatives are conceptually straightforward, but students often stumble on a few predictable pitfalls. The table below summarizes the most common mistakes alongside the correct approach.

Common Mistake	Why It's Wrong	Correct Approach
Differentiating both variables at once	When finding ∂f/∂x, treating y as a variable instead of a constant leads to an incorrect expression	Mentally replace every y with a specific number (like 5) to remind yourself it's constant, then differentiate
Forgetting constant terms don't vanish	In ∂/∂x of 3xy², students drop the y² instead of keeping it as a constant coefficient	y² is a constant when differentiating with respect to x; ∂/∂x of 3xy² = 3y², not 3
Mixing up the order in mixed partials	∂²f/∂y∂x means differentiate with respect to x first, then y — reading right to left	Read the denominator right to left: ∂y∂x → first ∂x, then ∂y. Subscript notation fₓᵧ reads left to right: first x, then y
Forgetting the chain rule	For terms like sin(xy), students forget the inner derivative	∂/∂x of sin(xy) = cos(xy) × y (chain rule multiplies by the derivative of the inner function with respect to x)

Common pitfalls when computing partial derivatives

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KEY TAKEAWAY

Think of each "other" variable like a fixed number baked into the function. If f(x, y) = x²y³ and you're finding ∂f/∂x, imagine y = 2 so the function looks like f(x) = 8x². Now it's obvious the derivative is 16x, which is the same as 2y³ × x when you put y back in. This mental substitution trick catches most errors.

SECTION 8

Connection to Advanced Topics

Partial derivatives are the building blocks for nearly everything else in multivariable calculus. Once you're comfortable computing them, you'll use them to construct more powerful tools. Here's a preview of where partial derivatives lead.

Concept	How It Uses Partial Derivatives	What It Does
Gradient (∇f)	Combines all first-order partials into a vector: ∇f = ⟨∂f/∂x, ∂f/∂y⟩	Points in the direction of steepest increase of f
Directional Derivative	Uses the gradient to find the rate of change in any direction, not just along axes	Generalizes the idea of a partial derivative beyond the x- and y-directions
Tangent Plane	Built from ∂f/∂x and ∂f/∂y evaluated at a point on a surface	Approximates the surface locally — the multivariable version of a tangent line
Second Derivative Test	Uses fₓₓ, fᵧᵧ, and fₓᵧ to classify critical points as maxima, minima, or saddle points	Extends the concavity test from single-variable calculus to surfaces
Partial Differential Equations	Equations like the heat equation relate partial derivatives of a function to each other	Models real-world phenomena: heat flow, wave propagation, fluid dynamics

Advanced topics that build on partial derivatives

Mastering partial derivatives now gives you the foundation for all of these topics. In particular, the gradient vector is the immediate next step — it packages your partial derivatives into a single object that captures how a function changes in every direction at once. If you're comfortable finding ∂f/∂x and ∂f/∂y, you're ready for that journey.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Explain in your own words what it means to compute ∂f/∂x for a function f(x, y). What role does y play during the differentiation, and why is the result generally still a function of both x and y?

Show answer

When computing ∂f/∂x, you differentiate f with respect to x while treating y as a constant — as if y were a fixed number like 3 or 7. The variable y plays the role of a constant coefficient during the differentiation process. The result is generally still a function of both x and y because, even though y was held constant during differentiation, the expression you obtain may still contain y. For example, if f(x, y) = x²y, then ∂f/∂x = 2xy, which depends on both variables.

Question 2: BASIC CALCULATION

Let f(x, y) = 4x³ − 5xy + 7y². Find ∂f/∂x and ∂f/∂y.

Show answer

To find ∂f/∂x, treat y as a constant: the derivative of 4x³ is 12x², the derivative of −5xy is −5y (since y is a constant coefficient), and the derivative of 7y² is 0 (pure constant with respect to x). So ∂f/∂x = 12x² − 5y. To find ∂f/∂y, treat x as a constant: the derivative of 4x³ is 0, the derivative of −5xy is −5x, and the derivative of 7y² is 14y. So ∂f/∂y = −5x + 14y.

Question 3: INTERMEDIATE

Let g(x, y) = eˣ·cos(y) + x²y³. Find all four second-order partial derivatives (gₓₓ, gᵧᵧ, gₓᵧ, gᵧₓ) and verify that the mixed partials are equal.

Show answer

First, find the first-order partials. gₓ = eˣ·cos(y) + 2xy³ and gᵧ = −eˣ·sin(y) + 3x²y². Now take second-order partials. gₓₓ = ∂/∂x(gₓ) = eˣ·cos(y) + 2y³. gᵧᵧ = ∂/∂y(gᵧ) = −eˣ·cos(y) + 6x²y. For the mixed partials: gₓᵧ = ∂/∂y(gₓ) = −eˣ·sin(y) + 6xy². gᵧₓ = ∂/∂x(gᵧ) = −eˣ·sin(y) + 6xy². Since gₓᵧ = gᵧₓ = −eˣ·sin(y) + 6xy², Clairaut's theorem is confirmed.

Question 4: APPLIED

The volume of a right circular cone is V(r, h) = (1/3)πr²h, where r is the radius and h is the height. Find ∂V/∂r and ∂V/∂h, then evaluate each at r = 3 cm and h = 10 cm. Interpret each result in physical terms.

Show answer

Treating h as a constant: ∂V/∂r = (1/3)π·2rh = (2/3)πrh. Treating r as a constant: ∂V/∂h = (1/3)πr². At r = 3 and h = 10: ∂V/∂r = (2/3)π(3)(10) = 20π ≈ 62.8 cm³/cm, and ∂V/∂h = (1/3)π(9) = 3π ≈ 9.42 cm³/cm. The first result means that if you increase the radius by 1 cm while keeping the height at 10 cm, the volume increases by approximately 62.8 cm³. The second means increasing the height by 1 cm while keeping the radius at 3 cm adds roughly 9.42 cm³ to the volume. The volume is much more sensitive to changes in radius than to changes in height.

Question 5: CRITICAL THINKING

Suppose f(x, y) = x²y + sin(xy). (a) Find ∂²f/∂x² and ∂²f/∂y². (b) Determine whether the function satisfies the equation ∂²f/∂x² + ∂²f/∂y² = 0 (Laplace's equation) at the point (0, 0). (c) Explain why Laplace's equation is significant — what would it mean physically if a temperature function T(x, y) satisfied this equation?

Show answer

First, fₓ = 2xy + y·cos(xy) (using chain rule on sin(xy)). Then fₓₓ = 2y − y²·sin(xy). For the y-direction: fᵧ = x² + x·cos(xy). Then fᵧᵧ = −x²·sin(xy). At (0, 0): fₓₓ(0,0) = 2(0) − 0²·sin(0) = 0, and fᵧᵧ(0,0) = −0²·sin(0) = 0. So fₓₓ + fᵧᵧ = 0 at the origin — yes, it satisfies Laplace's equation there. In general, however, fₓₓ + fᵧᵧ = 2y − y²sin(xy) − x²sin(xy) ≠ 0 everywhere. Physically, Laplace's equation describes a steady-state temperature distribution: if T(x, y) satisfies ∂²T/∂x² + ∂²T/∂y² = 0 everywhere, it means the temperature at each point is the average of nearby temperatures, and there are no internal heat sources or sinks — the system has reached thermal equilibrium.

SUMMARY

Lesson Summary

Summary

A partial derivative measures the rate of change of a multivariable function with respect to one variable while holding all other variables constant. You compute it by applying the same differentiation rules (power rule, product rule, chain rule) from single-variable calculus, treating every other variable as a constant. The notation ∂f/∂x uses the curly-d symbol to distinguish partial derivatives from ordinary ones. A function of two variables has two first-order partials, and each first-order partial can be differentiated again to produce second-order partials — both unmixed (fₓₓ, fᵧᵧ) and mixed (fₓᵧ, fᵧₓ).

Clairaut's theorem guarantees that for most functions you'll encounter, the mixed partial derivatives are equal regardless of the order of differentiation (fₓᵧ = fᵧₓ). Partial derivatives are the foundation for the gradient vector, tangent planes, the multivariable second derivative test, and partial differential equations — making them one of the most essential tools in all of higher mathematics and its applications.