Partial Differentiation

Updated 5 October 2026

Profit depends on price and on how many are sold, so it is a function of two variables. Partial differentiation is calculus for functions like these. Every rule of ordinary differentiation still applies. What changes is the geometry: the graph is a surface, and a surface has a different slope in each direction.

What is a function of two variables?

A function of two variables takes a pair of numbers (x, y) and gives back one number, written f(x, y). For $x^2+y^2$x² + y², the input (3, 4) gives 9 + 16 = 25. The input is a point in a plane, and the output is a height above it, so the graph is a surface, here a bowl with its lowest point at the origin. Ordinary calculus needs care here because a surface can slope differently in every direction. Level curves help: each is the slice at one fixed height, like a contour on a map, and this bowl at height 25 gives the circle of radius 5 through (3, 4). Some inputs must be left out, which changes the domain: $\sqrt{x-y}$√(x − y) needs $x\ge y$x ≥ y. The point (−3, 4) sits at the same height, because squaring removes the sign. See Functions of two variables and Surfaces in three dimensions.

Cut the bowl $z=x^2+y^2$z = x² + y² at height 4 and the slice is the circle $x^2+y^2=4$x² + y² = 4, which has radius 2. See Level curves.

Do limits still work with two variables?

Yes, but in two variables a limit must give the same value along every path into the point, not just from two sides. Take $f=2xy/\left(x^2+y^2\right)$f = 2xy / (x² + y²) near the origin. Along the x-axis, f is 0. Along the line y = x, f is $2x^2/2x^2$2x² / 2x², which is 1. Two paths give two values, so the limit does not exist. The value depends on the direction of approach: along y = mx it is $2m/\left(1+m^2\right)$2m / (1 + m²) at every distance. Agreement along two paths proves nothing, so showing a limit exists takes a different step, usually a bound. For $x^{2}y/\left(x^2+y^2\right)$x²y / (x² + y²), the size is at most |y|, which goes to 0, so that limit is 0. The line y = 2x makes the first function $\frac{4}{5}$4/5 throughout, with 2 × 2 = 4 on top and 1 + 4 = 5 underneath.

See Limits of functions of two variables.

Continuity means the same as it did for one variable: the limit at a point exists and equals the value of the function there. See Continuity in two variables.

What is a partial derivative?

A partial derivative is the rate of change of f in one direction, with the other variable held constant. To find $\frac{\partial f}{\partial x}$∂f/∂x, treat y as a constant and differentiate in x as usual; $\frac{\partial f}{\partial y}$∂f/∂y is the reverse. For $x^{2}y+5y^2$x²y + 5y², $\frac{\partial f}{\partial x}=2xy$∂f/∂x = 2xy, since y is a constant factor and $5y^2$5y² is a constant, and $\frac{\partial f}{\partial y}=x^2+10y$∂f/∂y = x² + 10y. At (1, 2) the x-slope is 2 × 1 × 2 = 4 and the y-slope is 1 + 10 × 2 = 21. Holding y fixed slices the surface parallel to the x-axis, and the slice is an ordinary curve whose gradient is $\frac{\partial f}{\partial x}$∂f/∂x. The curly $\partial$∂ replaces d to show that other variables are being held still, and a term with no x differentiates to 0 in x. The heights at (1, 2) and (2, 2) are 22 and 28, so that chord has gradient 28 − 22 = 6, above 4 because the slice curves upward. See The partial derivative.

No new rules are needed. See Computing partial derivatives.

Let $f=x^{2}y+y^3$f = x²y + y³. Then $\frac{\partial f}{\partial x}=2xy$∂f/∂x = 2xy, because y is a constant factor and $y^3$y³ is a constant. And $\frac{\partial f}{\partial y}=x^2+3y^2$∂f/∂y = x² + 3y², because $x^2$x² is a constant factor and $y^3$y³ differentiates to $3y^2$3y².

What about second derivatives?

Differentiating a partial derivative again gives a second partial derivative, and there are four: twice in x, twice in y, and the two mixed ones, x then y and y then x. Take $f=x^3{y}^2$f = x³y², whose first derivatives are $3x^2{y}^2$3x²y² in x and $2x^{3}y$2x³y in y. Differentiating again gives $\frac{{\partial }^{2}f}{\partial x^2}=6x{y}^2$∂²f/∂x² = 6xy² and $\frac{{\partial }^{2}f}{\partial y^2}=2x^3$∂²f/∂y² = 2x³, and both mixed derivatives are $6x^{2}y$6x²y. At (1, 2) those are 6 × 1 × 4 = 24, then 2, then 12 for each mixed one. The mixed pair agree by the mixed derivative theorem: when both are continuous, the order does not matter, so only three need working out. Together the three measure how the surface bends, and the test for a stationary point uses all of them. In the other order, $2x^{3}y$2x³y differentiated in x is $6x^{2}y$6x²y, and 6 × 1 × 2 = 12 again. See Higher partial derivatives.

Take $f=x^2{y}^3$f = x²y³. Then $\frac{\partial f}{\partial x}=2x{y}^3$∂f/∂x = 2xy³, and differentiating that in y gives $6x{y}^2$6xy². Also $\frac{\partial f}{\partial y}=3x^2{y}^2$∂f/∂y = 3x²y², and differentiating that in x gives $6x{y}^2$6xy². See The mixed derivative theorem.

How does the chain rule work here?

When x and y both depend on t, z = f(x, y) changes for two reasons at once, and $\frac{dz}{dt}$dz/dt adds one term per variable: the partial derivative of z with respect to it, times its rate of change. For z = xy with $x=t^2$x = t² and y = 3t, $\frac{\partial z}{\partial x}=y$∂z/∂x = y and $\frac{\partial z}{\partial y}=x$∂z/∂y = x, while the rates are 2t and 3, so $\frac{dz}{dt}=2$dz/dt = 2ty + 3x. At t = 1, where x = 1 and y = 3, that is 2 × 3 + 3 × 1 = 9. Over a short time each variable raises z by its own slope times its own step, and small rises add. If x and y each depend on s and t, then $\frac{\partial z}{\partial s}$∂z/∂s and $\frac{\partial z}{\partial t}$∂z/∂t each get one term through x and one through y. With one variable the formula is the ordinary chain rule. Substituting first agrees: $z=3t^3$z = 3t³ differentiates to $9t^2$9t², and 9 × 1 = 9.

$\frac{dz}{dt}=\left(\frac{\partial z}{\partial x}\right)\left(\frac{dx}{dt}\right)+\left(\frac{\partial z}{\partial y}\right)\left(\frac{dy}{dt}\right)$dz/dt = (∂z/∂x)(dx/dt) + (∂z/∂y)(dy/dt).

Draw the dependence as a tree, multiply along each branch, then add the branches. See The chain rule for partial derivatives.

What is the gradient vector?

The gradient vector $\nabla f$∇f has the two partial derivatives as its components, $\nabla f=\left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}\right)$∇f = (∂f/∂x, ∂f/∂y); it points in the direction of steepest ascent, and its length is the slope that way. For $x^2+y^2$x² + y², $\nabla f=\left(2x,2y\right)$∇f = (2x, 2y), so at (3, 4) it is (6, 8), pointing straight up the side of the bowl, with length 10, because 36 + 64 = 100. The slope along a unit vector u is $\nabla f\cdot u$∇f · u, which is $|\nabla f|\cos \theta$|∇f| cos θ for the angle $\theta$θ between them. That is largest at $\theta =0$θ = 0, so the gradient is the steepest way up, and it is 0 at a right angle, so the gradient crosses each level curve square on, here the circle of radius 5. A zero gradient means the surface is level, with no steepest direction. Along the unit vector $\left(\frac{3}{5},\frac{4}{5}\right)$(3/5, 4/5) the slope is $6\times \frac{3}{5}+8\times \frac{4}{5}=10$6 × 3/5 + 8 × 4/5 = 10, the full length. See The gradient vector.

This slope in a chosen direction is the directional derivative. See Directional derivatives.

The flat plane that rests on a surface at a point is the tangent plane. The two partial derivatives give its tilt in the x and y directions, and that fixes it. See Tangent planes.

How do you find maxima and minima on a surface?

At a maximum or minimum of a smooth surface, both partial derivatives are 0, so solve them together and then classify each point. For $f=x^2+2y^2-4x+4y$f = x² + 2y² − 4x + 4y, the partial derivatives 2x − 4 and 4y + 4 are both 0 only at (2, −1), where the height is 4 + 2 − 8 − 4 = −6. To classify, multiply the second derivatives in x and in y and subtract the square of the mixed one: 2 × 4 − 0 = 8, positive, and the x one is positive, so this is a minimum. A peak or a dip has a flat tangent plane, so both slopes vanish there. A negative result means a saddle: for $x^2-y^2$x² − y² at the origin it is 2 × (−2) − 0 = −4, rising one way and falling the other. Completing the square confirms it: f is ${\left(x-2\right)}^2+2{\left(y+1\right)}^2-6$(x − 2)² + 2(y + 1)² − 6, never below −6.

Because a saddle is possible, the second-derivative test in two variables uses all three second partial derivatives. See Stationary points of a surface.

What if the variables are constrained?

Under a constraint g(x, y) = 0, Lagrange multipliers find the largest or smallest value of f along that curve: set $\nabla f=\lambda \nabla g$∇f = λ∇g and solve the component equations together with the constraint. To maximize xy on the line x + y = 10, take g = x + y − 10. Then $\nabla f=\left(y,x\right)$∇f = (y, x) and $\nabla g=\left(1,1\right)$∇g = (1, 1), so $y=\lambda$y = λ and $x=\lambda$x = λ, which forces x = y = 5 and a maximum of 5 × 5 = 25. If $\nabla f$∇f crossed the curve at an angle, a step along the curve would still raise f, so at the best point the gradients are parallel. The multiplier means something too: $\lambda =5$λ = 5 is the rate at which the maximum grows as the 10 is raised. With two constraints each gets its own multiplier. Other points on the line give less, since 4 × 6 = 24 and 3 × 7 = 21.

Solve the component equations for x and y in terms of $\lambda$λ first, then substitute into the constraint. See Lagrange multipliers.

How do you integrate over a surface?

Integrating over a region of the plane means integrating twice, once in each variable, and gives the volume under the surface. Hold y constant and integrate in x to get the area of a cross-section, then integrate those areas in y. For xy over the rectangle with x from 0 to 2 and y from 0 to 3, the inner integral is $\frac{x^{2}y}{2}$x²y/2, which is 2y between the limits, and the outer integral of 2y is $y^2$y², which comes to 9. The volume is built from thin slices, each a cross-sectional area times a small thickness, just as an area is built from strips. A region that is not a rectangle changes the limits: over the triangle under y = x, the inner integral in y runs from 0 up to x. This rectangle's integral also splits into a product, because xy is x times y, and $2\times \frac{9}{2}=9$2 × 9/2 = 9.

In symbols, $A\left(y\right)=\int f\left(x,y\right)dx$A(y) = ∫ f(x, y) dx and $V=\int A\left(y\right)dy$V = ∫ A(y) dy. See The volume under a surface.

The mistakes worth naming

• Differentiating every variable at once. A term without your variable differentiates to 0.
• Expecting one number for the slope. The slope depends on direction, which is why the gradient is a vector.
• Believing a limit exists because two paths agree. The limit must be the same along every path in.
• Calling every stationary point a maximum or minimum. A saddle point makes both partial derivatives 0 and is neither.
• Substituting the constraint into f before forming the Lagrange equations. Take the gradients of the original f and g, then solve them together with the constraint.

The single-variable rules that carry over are in the rules of differentiation, the optimization ideas are in applications of differentiation, and the vector language is in vectors. Related guides: Applications of Integration and Polar Curves.