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Partial Derivatives

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Core Definition

Let z=f(x,y)z = f(x, y) be a function of two variables. The first-order partial derivative of ff with respect to xx at (x,y)(x, y) is defined by the difference quotient limit:

∂f∂x(x,y)=fx(x,y)=lim⁡h→0f(x+h,y)−f(x,y)h\frac{\partial f}{\partial x}(x, y) = f_x(x, y) = \lim_{h \to 0} \frac{f(x + h, y) - f(x, y)}{h}
Similarly, the first-order partial derivative of ff with respect to yy at (x,y)(x, y) is defined by:
∂f∂y(x,y)=fy(x,y)=lim⁡h→0f(x,y+h)−f(x,y)h\frac{\partial f}{\partial y}(x, y) = f_y(x, y) = \lim_{h \to 0} \frac{f(x, y + h) - f(x, y)}{h}
provided these limits exist.

Concept Explanation

The Core Operational Rule

In single-variable calculus, the derivative dfdx\frac{df}{dx} tracks the instantaneous rate of change of a curve y=f(x)y = f(x). In multivariable calculus, the input is multidimensional: a function z=f(x,y)z = f(x, y) defines a 2D surface hovering in 3D space R3\mathbb{R}^3. Because xx and yy can vary independently and along infinite combinations of trajectories, we isolate variation along one coordinate axis at a time:

The Fundamental Rule of Partial Differentiation:

  • To compute ∂f∂x\frac{\partial f}{\partial x}, differentiate with respect to xx treating yy as an absolute constant coefficient.
  • To compute ∂f∂y\frac{\partial f}{\partial y}, differentiate with respect to yy treating xx as an absolute constant coefficient.
Holding y = b fixed:    f(x, b) is purely a function of x  ===>  d/dx[f(x, b)] = f_x(x, b)
Holding x = a fixed:    f(a, y) is purely a function of y  ===>  d/dy[f(a, y)] = f_y(a, y)

Common Notational Conventions

The curly partial symbol "∂\partial" distinguishes partial derivatives from single-variable ordinary derivatives (dd):

  • With respect to xx:
    fx(x,y)=∂f∂x=∂z∂x=∂xf=D1f=Dxff_x(x, y) = \frac{\partial f}{\partial x} = \frac{\partial z}{\partial x} = \partial_x f = D_1 f = D_x f
  • With respect to yy:
    fy(x,y)=∂f∂y=∂z∂y=∂yf=D2f=Dyff_y(x, y) = \frac{\partial f}{\partial y} = \frac{\partial z}{\partial y} = \partial_y f = D_2 f = D_y f
  • Evaluated at a point (a,b)(a, b):
    fx(a,b)=∂f∂x∣(a,b)=∂z∂x∣(x=a,y=b)f_x(a, b) = \left.\frac{\partial f}{\partial x}\right|_{(a, b)} = \left.\frac{\partial z}{\partial x}\right|_{(x=a, y=b)}

Preservation of Single-Variable Calculus Rules

Because non-differentiated variables are treated as constants, every standard rule from single-variable differentiation applies directly:

  1. Constant Multiple Rule: ∂∂x[c(y)⋅g(x)]=c(y)⋅g′(x)\frac{\partial}{\partial x}[c(y) \cdot g(x)] = c(y) \cdot g'(x)
  2. Product Rule: ∂∂x[u(x,y)⋅v(x,y)]=∂u∂xv+u∂v∂x\frac{\partial}{\partial x}[u(x, y) \cdot v(x, y)] = \frac{\partial u}{\partial x} v + u \frac{\partial v}{\partial x}
  3. Quotient Rule: ∂∂x[u(x,y)v(x,y)]=∂u∂xv−u∂v∂xv2\frac{\partial}{\partial x}\left[\frac{u(x, y)}{v(x, y)}\right] = \frac{\frac{\partial u}{\partial x} v - u \frac{\partial v}{\partial x}}{v^2}
  4. Multivariable Chain Rule (Single Outer Function): ∂∂x[g(u(x,y))]=g′(u(x,y))⋅∂u∂x\frac{\partial}{\partial x}[g(u(x, y))] = g'(u(x, y)) \cdot \frac{\partial u}{\partial x}

Engineering Applications

  • Aerospace & Thermal Engineering: Fourier's law of heat conduction relates heat flux vector components to directional temperature gradients: qx=−k∂T∂xq_x = -k \frac{\partial T}{\partial x}.
  • Fluid Mechanics & Hydraulics: Viscous shear stress τ\tau in boundary-layer pipe flow depends on the transverse velocity gradient ∂v∂y\frac{\partial v}{\partial y}.
  • Electrical & Computer Engineering: In transistor semiconductor physics, gate leakage current variations ∂ID∂VGS\frac{\partial I_D}{\partial V_{GS}} define device transconductance.
  • Structural Mechanics: In plane stress analysis of airframes, normal strain rates ϵx=∂u∂x\epsilon_x = \frac{\partial u}{\partial x} determine elastic deformation along structural stringers.

Example Problems

Example 1: Differentiating with Polynomials and Radicals

Compute ∂f∂x\frac{\partial f}{\partial x} and ∂f∂y\frac{\partial f}{\partial y} for f(x,y)=x3y2−x2+y2f(x, y) = x^3 y^2 - \sqrt{x^2 + y^2}.

Solution

  1. Differentiate with respect to xx:

    ∂f∂x=∂∂x(x3y2)−∂∂x((x2+y2)1/2)\frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(x^3 y^2) - \frac{\partial}{\partial x}\left((x^2+y^2)^{1/2}\right)

    • Holding yy constant, ∂∂x(x3y2)=(3x2)(y2)=3x2y2\frac{\partial}{\partial x}(x^3 y^2) = (3x^2)(y^2) = 3x^2 y^2.
    • Using the chain rule on the radical:
      ∂∂x((x2+y2)1/2)=12(x2+y2)−1/2⋅∂∂x(x2+y2)=12x2+y2⋅2x=xx2+y2\frac{\partial}{\partial x}\left((x^2+y^2)^{1/2}\right) = \frac{1}{2}(x^2+y^2)^{-1/2} \cdot \frac{\partial}{\partial x}(x^2+y^2) = \frac{1}{2\sqrt{x^2+y^2}} \cdot 2x = \frac{x}{\sqrt{x^2+y^2}}
      ∂f∂x=3x2y2−xx2+y2\frac{\partial f}{\partial x} = 3x^2 y^2 - \frac{x}{\sqrt{x^2+y^2}}

  2. Differentiate with respect to yy:

    ∂f∂y=∂∂y(x3y2)−∂∂y((x2+y2)1/2)\frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(x^3 y^2) - \frac{\partial}{\partial y}\left((x^2+y^2)^{1/2}\right)

    • Holding xx constant, ∂∂y(x3y2)=(x3)(2y)=2x3y\frac{\partial}{\partial y}(x^3 y^2) = (x^3)(2y) = 2x^3 y.
    • Using the chain rule on the radical:
      ∂∂y((x2+y2)1/2)=12(x2+y2)−1/2⋅2y=yx2+y2\frac{\partial}{\partial y}\left((x^2+y^2)^{1/2}\right) = \frac{1}{2}(x^2+y^2)^{-1/2} \cdot 2y = \frac{y}{\sqrt{x^2+y^2}}
      ∂f∂y=2x3y−yx2+y2\frac{\partial f}{\partial y} = 2x^3 y - \frac{y}{\sqrt{x^2+y^2}}

Example 2: Differentiating Rational and Trigonometric Functions

Compute ∂f∂x\frac{\partial f}{\partial x} and ∂f∂y\frac{\partial f}{\partial y} for f(x,y)=xsin⁡(y)x2+1f(x, y) = \frac{x \sin(y)}{x^2 + 1}.

Solution

  1. Differentiate with respect to xx: Factor out the constant sin⁡(y)\sin(y):

    ∂f∂x=sin⁡(y)⋅∂∂x[xx2+1]\frac{\partial f}{\partial x} = \sin(y) \cdot \frac{\partial}{\partial x}\left[\frac{x}{x^2+1}\right]
    Apply the Quotient Rule u′v−uv′v2\frac{u'v - uv'}{v^2} where u=xu = x and v=x2+1v = x^2+1:
    ∂∂x[xx2+1]=(1)(x2+1)−(x)(2x)(x2+1)2=1−x2(x2+1)2\frac{\partial}{\partial x}\left[\frac{x}{x^2+1}\right] = \frac{(1)(x^2+1) - (x)(2x)}{(x^2+1)^2} = \frac{1 - x^2}{(x^2+1)^2}
    ∂f∂x=(1−x2)sin⁡(y)(x2+1)2\frac{\partial f}{\partial x} = \frac{(1 - x^2)\sin(y)}{(x^2+1)^2}

  2. Differentiate with respect to yy: Factor out the constant xx2+1\frac{x}{x^2+1}:

    ∂f∂y=xx2+1⋅∂∂y[sin⁡(y)]=xcos⁡(y)x2+1\frac{\partial f}{\partial y} = \frac{x}{x^2+1} \cdot \frac{\partial}{\partial y}[\sin(y)] = \frac{x\cos(y)}{x^2+1}

Connections & References