Partial Derivatives

Quick answer

A partial derivative measures how a function of several variables changes when one input moves and the others stay fixed. To compute ∂f/∂x, treat every other variable as a constant and differentiate with the ordinary rules. Geometrically, f_x(a, b) is the slope of the curve cut from the surface z = f(x, y) by the plane y = b. Differentiating again gives second partial derivatives, and for the functions met in practice the two mixed partials f_xy and f_yx are equal, which is Clairaut's theorem.

What you'll learn

Compute first partial derivatives with the usual rules
Interpret a partial derivative as the slope of a slice
Compute second partial derivatives, including mixed ones
Use partial derivatives as rates of change in context

One variable at a time

A function of two variables can change in many directions at once. The simplest way to measure change is to move in one coordinate direction and hold the other coordinate still. The partial derivative of $f$ with respect to $x$ is

f_x(a, b) = \frac{\partial f}{\partial x}(a, b) = \lim_{h \to 0} \frac{f(a + h, b) - f(a, b)}{h}

Only $x$ moves; $y$ stays at $b$ . So in practice, treat every other variable as a constant and differentiate as usual. The partial derivative $f_y$ is the same with the roles swapped.

Why a partial derivative is a slope

Fix $y = b$ and the surface $z = f(x, y)$ is cut by the vertical plane $y = b$ in a curve, $z = f(x, b)$ , a function of $x$ alone. Its ordinary derivative at $x = a$ is the limit above. So $f_x(a, b)$ is the slope of the slice of the surface through the point, taken in the $x$ direction. Likewise $f_y(a, b)$ is the slope of the slice in the $y$ direction.

slice: z = 2 − x²
tangent, slope −2

The slice y = 1 of z = 4 − x² − 2y², with its tangent line at x = 1

For $f(x, y) = 4 - x^2 - 2y^2$ : $f_x = -2x$ , so $f_x(1, 1) = -2$ , matching the tangent line in the figure. And $f_y = -4y$ , so $f_y(1, 1) = -4$ : the slice in the $y$ direction falls twice as steeply.

Worked examples

Common mistakes

Practice problems

Find $f_x$ and $f_y$ for $f(x, y) = 3x^2y - 5y^3$ , and evaluate them at $(1, 2)$ .

Answer

$f_x = 6xy$ , $f_y = 3x^2 - 15y^2$ ; at $(1, 2)$ : $12$ and $-57$

Full solution

Hold $y$ fixed for $f_x$ and $x$ fixed for $f_y$ . Then $6 \cdot 1 \cdot 2 = 12$ and $3 - 60 = -57$ .
Find $f_x$ for $f(x, y) = e^{xy}$ , and evaluate it at $(0, 3)$ .

Answer

$f_x = ye^{xy}$ ; $3$

Full solution

The chain rule multiplies by the $x$ -derivative of $xy$ , which is $y$ . At $(0, 3)$ : $3e^0 = 3$ .
Find both partial derivatives of $f(x, y) = \tfrac{x}{y}$ at $(4, 2)$ .

Answer

$f_x = \tfrac{1}{2}$ and $f_y = -1$

Full solution

$f_x = \tfrac{1}{y}$ and $f_y = -\tfrac{x}{y^2}$ . At $(4, 2)$ : $\tfrac{1}{2}$ and $-\tfrac{4}{4}$ .
Find $f_x$ for $f(x, y) = \ln(x^2 + y^2)$ at $(1, 1)$ .

Answer

$1$

Full solution

$f_x = \tfrac{2x}{x^2 + y^2}$ , which is $\tfrac{2}{2}$ at $(1, 1)$ .
Find $f_{xy}$ for $f(x, y) = x^2\sin y$ .

Answer

$2x\cos y$

Full solution

$f_x = 2x\sin y$ , then differentiate in $y$ . Starting with $f_y = x^2\cos y$ and differentiating in $x$ gives the same.
Find $f_{xx}$ and $f_{yy}$ for $f(x, y) = x^4 - 3x^2y + y^2$ .

Answer

$f_{xx} = 12x^2 - 6y$ and $f_{yy} = 2$

Full solution

$f_x = 4x^3 - 6xy$ and $f_y = -3x^2 + 2y$ . Differentiate each again in the same variable.
For $f(x, y) = 4 - x^2 - 2y^2$ , find $f_y(1, 1)$ and say what it means.

Answer

$-4$ ; the slice $x = 1$ falls with slope $-4$ at $y = 1$ .

Full solution

$f_y = -4y$ . In the plane $x = 1$ the surface is $z = 3 - 2y^2$ , whose slope at $y = 1$ is $-4$ .
For $V(r, h) = \pi r^2h$ , find $V_r$ and $V_h$ at $r = 3$ , $h = 10$ .

Answer

$V_r = 60\pi$ and $V_h = 9\pi$

Full solution

$V_r = 2\pi rh$ and $V_h = \pi r^2$ . Widening the cylinder changes its volume more than lengthening it, at this size.
Find $f_z$ for $f(x, y, z) = x^2 + yz^2$ at $(1, 2, 3)$ .

Answer

$12$

Full solution

$f_z = 2yz = 2 \cdot 2 \cdot 3$ .
A student finds $f_x = 2x \cdot 3y^2$ for $f(x, y) = x^2y^3$ . What went wrong?

Hint

What happens to $y^3$ when $y$ is held constant?

Answer

The factor $y^3$ is a constant for $f_x$ and should not be differentiated. $f_x = 2xy^3$ .

Full solution

The student differentiated both factors, as if taking a total derivative. For $f_x$ , only $x$ varies, so $y^3$ rides along as a coefficient.

Frequently asked questions

How do you take a partial derivative?

Differentiate with respect to one variable while treating every other variable as a constant. All the rules of single-variable calculus apply.

What does a partial derivative mean geometrically?

f_x(a, b) is the slope, in the x direction, of the curve where the surface z = f(x, y) meets the vertical plane y = b.

What notations are used for partial derivatives?

∂f/∂x, f_x and ∂z/∂x all mean the partial derivative with respect to x. The curly ∂ signals that other variables are being held fixed.

What are mixed partial derivatives?

Second derivatives taken with respect to two different variables, such as f_xy, which differentiates first in x and then in y.

Does the order of differentiation matter?

Not when the second partial derivatives are continuous, which covers nearly every function met in practice. Clairaut's theorem then gives f_xy = f_yx.

What to learn next

Tangent Planes and Linear ApproximationBuild the plane that touches a surface z = f(x, y) at a point from the two partial derivatives, use it to approximate f nearby, and estimate errors with differentials.

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