Partial Derivatives and Total Differentials

Definition of Partial Derivatives

Partial Derivative

Let z=f(x,y)z = f(x, y) be defined on a neighborhood of (x0,y0)(x_0, y_0). Fix y=y0y = y_0 and give xx an increment Δx\Delta x at x0x_0. If the limit

limΔx0f(x0+Δx,y0)f(x0,y0)Δx\lim_{\Delta x \to 0} \frac{f(x_0 + \Delta x, y_0) - f(x_0, y_0)}{\Delta x}

exists, it is called the partial derivative of ff with respect to xx at (x0,y0)(x_0, y_0), written

fx(x0,y0),zx(x0,y0),fx(x0,y0),zx(x0,y0)f_x(x_0, y_0), \quad \frac{\partial z}{\partial x}\bigg|_{(x_0, y_0)}, \quad \frac{\partial f}{\partial x}\bigg|_{(x_0, y_0)}, \quad z_x\big|_{(x_0, y_0)}

Similarly, the partial derivative with respect to yy is

fy(x0,y0)=limΔy0f(x0,y0+Δy)f(x0,y0)Δyf_y(x_0, y_0) = \lim_{\Delta y \to 0} \frac{f(x_0, y_0 + \Delta y) - f(x_0, y_0)}{\Delta y}

If the partial derivative with respect to xx exists at every point of a region DD, it becomes a function of (x,y)(x,y), called the partial derivative function (often also just the partial derivative), written fx(x,y)f_x(x,y) or zx\frac{\partial z}{\partial x}.

Geometric Meaning

Geometric meaning of partial derivatives
  • fx(x0,y0)f_x(x_0, y_0): the slope of the tangent line, at (x0,y0,f(x0,y0))(x_0, y_0, f(x_0,y_0)), of the curve {z=f(x,y0)y=y0\begin{cases} z = f(x, y_0) \\ y = y_0 \end{cases} obtained by cutting the surface z=f(x,y)z = f(x,y) with the plane y=y0y = y_0, measured against the xx-axis.
  • fy(x0,y0)f_y(x_0, y_0): the slope of the tangent line of the curve cut by the plane x=x0x = x_0, measured against the yy-axis.

Partial Derivatives and Continuity

Higher-Order Partial Derivatives

Mixed Partial Derivatives

The second partial derivatives of z=f(x,y)z = f(x,y) are:

2zx2,2zy2,2zxy=x(zy),2zyx=y(zx)\frac{\partial^2 z}{\partial x^2}, \quad \frac{\partial^2 z}{\partial y^2}, \quad \frac{\partial^2 z}{\partial x\partial y} = \frac{\partial}{\partial x}\left(\frac{\partial z}{\partial y}\right), \quad \frac{\partial^2 z}{\partial y\partial x} = \frac{\partial}{\partial y}\left(\frac{\partial z}{\partial x}\right)

The mixed ones 2zxy\frac{\partial^2 z}{\partial x\partial y} and 2zyx\frac{\partial^2 z}{\partial y\partial x} are called mixed partial derivatives.

定理1

Clairaut’s theorem: if the mixed partials 2zxy\frac{\partial^2 z}{\partial x\partial y} and 2zyx\frac{\partial^2 z}{\partial y\partial x} are both continuous on a region, then

2zxy=2zyx\frac{\partial^2 z}{\partial x\partial y} = \frac{\partial^2 z}{\partial y\partial x}

Chain Rule

定理2

If z=f(x,y)z = f(x,y), where x=x(t)x = x(t) and y=y(t)y = y(t) are differentiable, then

dzdt=zxdxdt+zydydt\frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt}

More generally, if x=x(s,t)x = x(s,t) and y=y(s,t)y = y(s,t),

zs=zxxs+zyys,zt=zxxt+zyyt\frac{\partial z}{\partial s} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial s}, \qquad \frac{\partial z}{\partial t} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial t}

Total Differential and Differentiability

Total Differential

Let z=f(x,y)z = f(x, y) be defined on a neighborhood of (x0,y0)(x_0, y_0). If the total increment

Δz=f(x0+Δx,y0+Δy)f(x0,y0)\Delta z = f(x_0 + \Delta x, y_0 + \Delta y) - f(x_0, y_0)

can be written as

Δz=AΔx+BΔy+o(ρ)\Delta z = A\Delta x + B\Delta y + o(\rho)

where A,BA, B do not depend on Δx,Δy\Delta x, \Delta y and ρ=(Δx)2+(Δy)2\rho = \sqrt{(\Delta x)^2 + (\Delta y)^2}, then ff is differentiable at (x0,y0)(x_0, y_0), and AΔx+BΔyA\Delta x + B\Delta y is its total differential:

dz=AΔx+BΔyordf=AΔx+BΔydz = A\Delta x + B\Delta y \quad \text{or} \quad df = A\Delta x + B\Delta y

The total differential is the linear principal part of the total increment: as ρ0\rho \to 0, the remainder Δzdz=o(ρ)\Delta z - dz = o(\rho) is an infinitesimal of higher order than ρ\rho.

Necessary Condition for Differentiability

定理3

If z=f(x,y)z = f(x, y) is differentiable at (x,y)(x, y), then both partial derivatives exist, and

A=zx,B=zyA = \frac{\partial z}{\partial x}, \quad B = \frac{\partial z}{\partial y}

so the total differential is

dz=zxdx+zydydz = \frac{\partial z}{\partial x} dx + \frac{\partial z}{\partial y} dy

For a three-variable function u=f(x,y,z)u = f(x, y, z):

du=uxdx+uydy+uzdzdu = \frac{\partial u}{\partial x} dx + \frac{\partial u}{\partial y} dy + \frac{\partial u}{\partial z} dz

Sufficient Condition for Differentiability

定理4

If the partial derivatives zx\frac{\partial z}{\partial x} and zy\frac{\partial z}{\partial y} exist in a neighborhood of (x0,y0)(x_0, y_0) and are continuous at (x0,y0)(x_0, y_0), then ff is differentiable at (x0,y0)(x_0, y_0).

Relation: Differentiable / Differentiable-Partials / Continuous

Worked Examples

Example 1: Computing Partial Derivatives

Find the partial derivatives of f(x,y)=x2y+y3f(x, y) = x^2 y + y^3 at (1,2)(1, 2).

Answer and Explanation(3 个标签)
partial derivativecomputationpolynomial

Detailed steps:

  1. With respect to xx (treat yy as a constant): fx=2xyf_x = 2xy
  2. With respect to yy (treat xx as a constant): fy=x2+3y2f_y = x^2 + 3y^2
  3. Substitute (1,2)(1,2): fx(1,2)=2×1×2=4f_x(1,2) = 2\times1\times2 = 4, fy(1,2)=1+3×4=13f_y(1,2) = 1 + 3\times4 = 13

Answer: fx(1,2)=4f_x(1,2) = 4, fy(1,2)=13f_y(1,2) = 13.

Example 2: Partial Derivatives of a Piecewise Function

Find the partial derivatives of f(x,y)={x2yx2+y2,(x,y)(0,0)0,(x,y)=(0,0)f(x,y) = \begin{cases} \frac{x^2 y}{x^2+y^2}, & (x,y)\neq(0,0) \\ 0, & (x,y)=(0,0) \end{cases} at (0,0)(0,0).

Answer and Explanation(3 个标签)
partial derivativepiecewise functiondefinition method

Detailed steps:

  1. Using the definition for fx(0,0)f_x(0,0): fx(0,0)=limΔx0f(Δx,0)f(0,0)Δx=limΔx000Δx=0f_x(0,0) = \lim_{\Delta x\to0} \frac{f(\Delta x,0)-f(0,0)}{\Delta x} = \lim_{\Delta x\to0} \frac{0-0}{\Delta x} = 0
  2. Similarly fy(0,0)=0f_y(0,0) = 0
  3. Note: this function is actually continuous at the origin (its limit is 00, equal to f(0,0)f(0,0)), unlike the earlier counterexample.

Answer: fx(0,0)=0f_x(0,0) = 0, fy(0,0)=0f_y(0,0) = 0.

Example 3: Total Differential and Linear Approximation

Find the total differential of f(x,y)=x2yf(x, y) = x^2 y at (1,2)(1, 2), and use it to approximate f(1.01,1.98)f(1.01, 1.98).

Answer and Explanation(3 个标签)
total differentiallinear approximationapplication

Detailed steps:

  1. fx=2xyf_x = 2xy, so fx(1,2)=4f_x(1,2) = 4
  2. fy=x2f_y = x^2, so fy(1,2)=1f_y(1,2) = 1
  3. The total differential at (1,2)(1,2) is dz=4dx+1dydz = 4\,dx + 1\,dy
  4. With dx=0.01dx = 0.01, dy=0.02dy = -0.02: Δzdz=4(0.01)+1(0.02)=0.02\Delta z \approx dz = 4(0.01) + 1(-0.02) = 0.02
  5. f(1,2)=2f(1,2) = 2, so f(1.01,1.98)2+0.02=2.02f(1.01, 1.98) \approx 2 + 0.02 = 2.02

Answer: dz=4dx+dydz = 4dx + dy; the approximation is 2.022.02 (exact value 2.022.02 to two decimals).


Summary

Symbols Used in This Article

SymbolTypeReading/ExplanationMeaning in This Article
fx,fyf_x, f_ypartial derivativesf sub x, f sub yPartial derivatives with respect to xx and yy
zx\frac{\partial z}{\partial x}partial derivativepartial of z with respect to xPartial derivative in Leibniz notation
dzdztotal differentiald zThe total differential of zz
A,BA, BconstantsA, BCoefficients of the linear principal part
ρ\rhoGreek letterrho(Δx)2+(Δy)2\sqrt{(\Delta x)^2 + (\Delta y)^2}
Δz\Delta zincrementdelta zThe total increment of ff

English–Chinese Glossary

English termChinese termPhoneticExplanation
partial derivative偏导数/ˈpɑːʃəl dɪˈrɪvətɪv/The derivative with respect to one variable, others fixed
mixed partial derivative混合偏导数/mɪkst ˈpɑːʃəl dɪˈrɪvətɪv/A second partial with two different variables
total differential全微分/ˈtəʊtl ˌdɪfəˈrenʃəl/The linear principal part of the total increment
differentiability可微性/ˌdɪfəˌrenʃiəˈbɪləti/The property of having a total differential
chain rule链式法则/tʃeɪn ruːl/The rule for differentiating compositions
linear principal part线性主部/ˈlɪniə ˈprɪnsəpəl pɑːt/The linear term dominating the increment
Clairaut’s theorem克莱罗定理/ˈkleərəʊz ˈθɪərəm/Equality of mixed partials under continuity