Difficulty: Intermediate | Prerequisites: single-variable differentiation (Calc 1), familiarity with multivariable functions z = f(x, y).
Up to this point in calculus, every function you have differentiated has had one input. Partial derivatives extend differentiation to functions of two or more variables, letting you measure how the output changes when you nudge just one input while holding the others fixed. This is the gateway to gradients, directional derivatives, and optimisation in multiple dimensions. If you are comfortable taking derivatives of single-variable functions (including product rule, chain rule, and trig derivatives), you have everything you need to start here.
A partial derivative measures the rate of change of a multivariable function with respect to one variable, treating every other variable as a constant. You compute them with the same rules you already know from Calc 1. Second-order partials follow the same idea, and Clairaut's theorem tells you the mixed partials f_xy and f_yx are equal (provided they are continuous).
Partial derivative with respect to x (f_x or ∂f/∂x)
The derivative of f(x, y) taken while treating y as a constant. Formally: lim as h → 0 of [f(x + h, y) − f(x, y)] / h.
Think of it as: "How fast does the output change if I move only in the x-direction?"
Partial derivative with respect to y (f_y or ∂f/∂y)
The derivative of f(x, y) taken while treating x as a constant. Formally: lim as h → 0 of [f(x, y + h) − f(x, y)] / h.
In simple terms, this means the same idea as f_x, but you are sliding along the y-axis instead.
Second-order partial derivative (f_xx, f_yy, f_xy, f_yx)
The partial derivative of a partial derivative. For example, f_xy means: first differentiate with respect to x, then differentiate that result with respect to y.
Think of it as: "How does the rate of change itself change when I nudge a second variable?"
Clairaut's theorem (equality of mixed partials)
If f_xy and f_yx are both continuous, then f_xy = f_yx. The order of differentiation does not matter.
In simple terms, this means you can take mixed second derivatives in whichever order is easier, and you will get the same answer.
Contour plot (level curves)
A map of a surface z = f(x, y) drawn in the xy-plane, where each curve connects points that share the same z-value. Moving across tightly packed contours means the function is changing rapidly in that direction.
Think of it as: a topographic map. Closely spaced lines mean a steep slope.
For a function z = f(x, y), the two first-order partial derivatives are defined by limits:
f_x(x,y) = \lim_{h \to 0} \frac{f(x+h,\, y) - f(x,\, y)}{h}f_y(x,y) = \lim_{h \to 0} \frac{f(x,\, y+h) - f(x,\, y)}{h}You will rarely need to evaluate these limits directly. In practice, you use standard differentiation rules, treating the other variable as a constant.
The technique is straightforward: to find f_x, treat y as a constant and differentiate with respect to x using every rule you already know (power rule, product rule, chain rule, trig rules). To find f_y, do the reverse.
Example 1: Polynomial
f(x, y) = 3x² + 4y² + x²y
f_x = 6x + 2xy (differentiate each term with respect to x, treating y as a constant)
f_y = 8y + x² (differentiate each term with respect to y, treating x as a constant)
Example 2: Product with trig
f(x, y) = x² cos(xy)
This requires the product rule. Let u = x² and v = cos(xy).
f_x = 2x cos(xy) + x²(−y sin(xy)) = 2x cos(xy) − x²y sin(xy)
f_y = x²(−x sin(xy)) = −x³ sin(xy)
Example 3: Mixed polynomial
f(x, y) = 7xy + 2x²y − y
f_x = 7y + 4xy
f_y = 7x + 2x² − 1
Example 4: Exponential with product rule
f(x, y) = (x + y)e^(3x − 2y)
Using the product rule on (x + y) and e^(3x − 2y):
f_x = 1 · e^(3x − 2y) + (x + y) · 3e^(3x − 2y) = e^(3x − 2y)[1 + 3(x + y)]
Evaluating at (1, 1): f_x(1, 1) = e^(3 − 2)[1 + 3(2)] = 7e.
The slope of the surface in the x-direction at the point (1, 1, f(1, 1)) is 7e.
Example 5: Fundamental Theorem of Calculus
f(x, y) = ∫ from y to x of sin⁵(t) dt
By the FTC, differentiating with respect to the upper limit: f_x = sin⁵(x).
Differentiating with respect to the lower limit (with the sign flip): f_y = −sin⁵(y).
Example 6: FTC with exponential integrand
f(x, y) = ∫ from y to x of 4e^(t²) dt
f_y = −4e^(y²). Evaluating at (0, 0): f_y(0, 0) = −4.
Once you have f_x and f_y, you can differentiate again to get four second-order partials:
f_xx = ∂/∂x (f_x), differentiate f_x with respect to x again
f_yy = ∂/∂y (f_y), differentiate f_y with respect to y again
f_xy = ∂/∂y (f_x), differentiate f_x with respect to y
f_yx = ∂/∂x (f_y), differentiate f_y with respect to x
Clairaut's theorem states that if f_xy and f_yx are both continuous, then f_xy = f_yx. In every problem you will meet in this course, the mixed partials will be equal. This means you can pick whichever order of differentiation is simpler.
Worked example
f(x, y) = sin(x + 2y)
f_x = cos(x + 2y)
f_y = 2 cos(x + 2y)
f_xx = −sin(x + 2y)
f_yy = −4 sin(x + 2y)
f_xy = −2 sin(x + 2y)
Note that f_yx = ∂/∂x [2 cos(x + 2y)] = −2 sin(x + 2y), confirming Clairaut's theorem.
Another worked example
f(x, y) = 2x³y tan(3x + 1)
f_x = 6x²y tan(3x + 1) + 2x³y · 3 sec²(3x + 1) = 6x²y tan(3x + 1) + 6x³y sec²(3x + 1)
f_xy = 6x² tan(3x + 1) + 6x³ sec²(3x + 1)
Here the y-differentiation step is simple because y appears as a single linear factor in f_x.
A contour plot (level-curve map) lets you determine the sign of f_x and f_y without computing anything.
Sign of f_x: Move to the right from the point in question. If you cross contours toward higher values of f, then f_x > 0. If toward lower values, f_x < 0.
Sign of f_y: Move upward. Same logic: higher contours mean f_y > 0, lower contours mean f_y < 0.
The spacing of the contours tells you the magnitude. Tightly packed contours in a given direction mean a large partial derivative in that direction (the function is changing quickly). Widely spaced contours mean a small partial derivative (the function is nearly flat).
Example from the lecture notes
A contour plot showed values decreasing from left to right and increasing from bottom to top at point P. Conclusion: f_x < 0 (moving right decreases f) and f_y > 0 (moving up increases f).
f_x = \frac{\partial f}{\partial x} = \lim_{h \to 0} \frac{f(x+h, y) - f(x, y)}{h}f_y = \frac{\partial f}{\partial y} = \lim_{h \to 0} \frac{f(x, y+h) - f(x, y)}{h}f_{xy} = \frac{\partial^2 f}{\partial y \, \partial x} = \frac{\partial}{\partial y}\!\left(\frac{\partial f}{\partial x}\right)Clairaut's theorem: f_xy = f_yx (when both mixed partials are continuous).
FTC connection: If f(x, y) = ∫ from y to x of g(t) dt, then f_x = g(x) and f_y = −g(y).
Partial derivatives show up anywhere a quantity depends on more than one input. In thermodynamics, the pressure of a gas depends on both temperature and volume; ∂P/∂T tells engineers how pressure responds to a temperature change at constant volume. In economics, a firm's profit depends on price and quantity, and partial derivatives give the marginal effect of changing one while holding the other fixed. In machine learning, gradient descent relies on partial derivatives of a loss function with respect to each model parameter to find the combination that minimises error.
Students often forget to apply the chain rule when the "constant" variable appears inside a composed function. In f(x, y) = x² cos(xy), computing f_x requires the product rule and the chain rule on cos(xy), because x appears in the argument of cos as well.
Students sometimes confuse the notation f_xy with "differentiate with respect to x then y" when it actually means "differentiate with respect to x first (inner), then with respect to y (outer)." The subscript order reads left to right: x first, y second.
When using the Fundamental Theorem of Calculus, students forget the sign flip for the lower limit. If f(x, y) = ∫ from y to x of g(t) dt, then f_y = −g(y), not +g(y).
Students occasionally treat ∂f/∂x as a fraction and try to cancel the ∂ symbols. The ∂ notation is not a fraction; it is a single operator symbol.
⚠️ Expect problems that combine the product rule or chain rule with partial differentiation. A function like x² cos(xy) tests whether you can juggle both at once.
⚠️ Clairaut's theorem is frequently tested by asking you to verify f_xy = f_yx for a given function, or by asking you to compute a mixed partial in whichever order is easier.
⚠️ FTC-based partial derivative problems (integrals with variable limits) appear regularly. Know the sign convention for the lower vs. upper limit.
⚠️ Reading the sign of partial derivatives from a contour plot is a common conceptual question on exams. Practice interpreting level curves before the test.
True or False: To compute f_x, you treat x as a constant and differentiate with respect to y.
Answer: False. To compute f_x, you treat y as a constant and differentiate with respect to x.
Fill in the blank: If f(x, y) = 5x³y², then f_y = ______.
Answer: 10x³y.
True or False: Clairaut's theorem guarantees f_xy = f_yx for all functions.
Answer: False. It requires that both mixed partials are continuous.
Fill in the blank: If f(x, y) = ∫ from y to x of e^(t³) dt, then f_x = ______.
Answer: e^(x³).
True or False: On a contour plot, if moving to the right takes you from the contour labelled 10 to the contour labelled 6, then f_x < 0 at that point.
Answer: True.
Q: Let f(x, y) = x³y + 2xy⁴. Find f_x and f_y.
A: f_x = 3x²y + 2y⁴. f_y = x³ + 8xy³.
Q: Let f(x, y) = e^(xy). Compute f_xy.
A: f_x = ye^(xy). Then f_xy = e^(xy) + xye^(xy) = e^(xy)(1 + xy).
Q: Verify Clairaut's theorem for f(x, y) = x² sin(y).
A: f_x = 2x sin(y), so f_xy = 2x cos(y). f_y = x² cos(y), so f_yx = 2x cos(y). They are equal.
Q: If f(x, y) = ∫ from y to x of cos(t²) dt, find f_x and f_y and evaluate f_y(0, 0).
A: f_x = cos(x²). f_y = −cos(y²). f_y(0, 0) = −cos(0) = −1.
Q: A contour plot of f(x, y) shows contour values increasing as you move left and downward from a point P. What are the signs of f_x and f_y at P?
A: Moving right (positive x-direction) decreases f, so f_x < 0. Moving up (positive y-direction) decreases f, so f_y < 0.
Partial derivatives are the building blocks for the gradient vector ∇f = (f_x, f_y), which points in the direction of steepest ascent and underpins directional derivatives and optimisation (finding maxima, minima, and saddle points via the second derivative test in two variables). They also feed into the chain rule for multivariable functions, which you will need when variables themselves depend on other parameters. If you go on to vector calculus, divergence and curl are both defined in terms of partial derivatives.
partial derivatives, partial differentiation, ∂f/∂x, ∂f/∂y, f_x, f_y, multivariable calculus, functions of two variables, Clairaut's theorem, equality of mixed partials, second-order partial derivatives, f_xx, f_yy, f_xy, f_yx, mixed partials, contour plot, level curves, slope in x-direction, slope in y-direction, FTC and partial derivatives, Calc 2, M408D, University of Texas, integral calculus, gradient prerequisites