Learning goals

What you should understand

  • Explain a partial derivative as the slope of a coordinate-direction slice through a surface.
  • Differentiate with respect to x, y, or z while holding the other independent variables constant.
  • Evaluate a partial derivative at a point only after finding the symbolic derivative and checking the domain.
  • Distinguish partial differentiation from total and implicit differentiation.
Before you begin

You should know single-variable power, product, and chain rules, function notation, coordinate points, and the idea of viewing a surface through a fixed-coordinate slice.

Hold one coordinate fixed and differentiate the slice

Each definition is an ordinary difference quotient along one coordinate direction. The variable not receiving h stays fixed, so the surface is replaced locally by a one-variable curve.

Change x while y stays fixedfx(x,y)=lim⁡h→0f(x+h,y)−f(x,y)hf_x(x,y)=\lim_{h\to0}\frac{f(x+h,y)-f(x,y)}{h}
Change y while x stays fixedfy(x,y)=lim⁡h→0f(x,y+h)−f(x,y)hf_y(x,y)=\lim_{h\to0}\frac{f(x,y+h)-f(x,y)}{h}
  • The relevant limit must exist at the point for that partial derivative to exist.
  • The point must lie in the domain, with enough nearby points in the selected coordinate direction.
  • Existing partial derivatives at a point do not alone guarantee continuity or total differentiability there.

Reusable process

Use the same fixed-variable decision every time

  1. 1
    Name the active variable

    Write the requested symbol, such as partial f over partial x, before doing algebra.

    fx=∂f∂xf_x=\frac{\partial f}{\partial x}
  2. 2
    Freeze the other independent variables

    Treat their powers and functions as constants during this derivative.

  3. 3
    Apply single-variable rules

    Use the power, product, quotient, and chain rules exactly as you would for a one-variable slice.

  4. 4
    Simplify without losing restrictions

    Keep exclusions from logarithms, roots, and denominators beside the result.

  5. 5
    Evaluate and verify

    Substitute coordinates last, then compare a small coordinate-direction difference quotient at a valid point.

Coordinate slice

Freeze y, then watch the x-slope change

For f(x,y) = x squared + 2xy + y squared, choose a fixed y-value. Each choice creates a one-variable parabola in x whose derivative is 2x + 2y.

f(x,y)=x2+2xy+y2=(x+y)2f(x,y)=x^2+2xy+y^2=(x+y)^2
1
Fixed coordinatey = 1
Slicef(x, 1) = (x + 1)²
x-direction slope2x + 2(1)
Parabolic coordinate sliceThe fixed y value is 1. The displayed parabola is the resulting one-variable slice.

If the control is unavailable, set y equal to any constant c. The slice f(x,c) has derivative 2x + 2c, so substituting c into 2x + 2y gives the same slope.

Check the reasoning

With y fixed at c, the slice is f(x,c) = x squared + 2cx + c squared. Its slope is 2x + 2c, which is exactly f sub x evaluated at y = c. Changing c shifts the slice and its slope without changing which variable is differentiated.

Worked examples

Build from recognition to verification

Polynomial surface and point evaluation

Hold y fixed, differentiate with respect to x, and substitute the point only after the symbolic step.

f(x,y)=x2y+3y2,fx(1,−2)f(x,y)=x^2y+3y^2,\qquad f_x(1,-2)
  1. Freeze y: The factors y and 3y squared act as constants in the x direction.
    ∂∂x(3y2)=0\frac{\partial}{\partial x}(3y^2)=0
  2. Differentiate: Apply the x power rule and keep the coefficient y.
    fx(x,y)=2xyf_x(x,y)=2xy
  3. Evaluate: Insert x = 1 and y = -2.
    fx(1,−2)=2(1)(−2)=−4f_x(1,-2)=2(1)(-2)=-4

Answer

fx(1,−2)=−4f_x(1,-2)=-4

Check: With y fixed at -2, f(x,-2) = -2x squared + 12. Its ordinary derivative is -4x, which equals -4 at x = 1.

Exponential chain rule

The exponent contains both variables, but x is constant during the y-partial derivative.

g(x,y)=exy,gy(2,0)g(x,y)=e^{xy},\qquad g_y(2,0)
  1. Differentiate the exponent: With x fixed, the derivative of xy with respect to y is x.
    ∂∂y(xy)=x\frac{\partial}{\partial y}(xy)=x
  2. Apply the chain rule: Multiply e to the xy by the inner derivative.
    gy(x,y)=xexyg_y(x,y)=xe^{xy}
  3. Evaluate: At (2,0), the exponential factor is one.
    gy(2,0)=2e0=2g_y(2,0)=2e^0=2

Answer

gy(2,0)=2g_y(2,0)=2

Check: Fixing x at 2 gives the one-variable slice e to the 2y, whose derivative at y = 0 is 2.

Logarithmic surface with a domain condition

The logarithm is defined whenever x squared plus y squared is positive, so the origin is excluded.

h(x,y)=ln⁡(x2+y2),hx(3,4)h(x,y)=\ln(x^2+y^2),\qquad h_x(3,4)
  1. Differentiate the outer function: The derivative of ln u is one over u times the derivative of u.
    hx=1x2+y2∂∂x(x2+y2)h_x=\frac{1}{x^2+y^2}\frac{\partial}{\partial x}(x^2+y^2)
  2. Differentiate the inside: The y squared term is constant in the x direction.
    hx(x,y)=2xx2+y2h_x(x,y)=\frac{2x}{x^2+y^2}
  3. Evaluate: The point is in the domain because 3 squared plus 4 squared is positive.
    hx(3,4)=625h_x(3,4)=\frac6{25}

Answer

hx(3,4)=625h_x(3,4)=\frac6{25}

Check: Fix y at 4 and differentiate ln(x squared + 16) as a one-variable function. Its derivative is 2x/(x squared + 16), giving 6/25 at x = 3.

Three-variable temperature model

Only z changes, so every factor containing only x or y is constant.

T(x,y,z)=x2y+yz3,Tz(1,−2,2)T(x,y,z)=x^2y+yz^3,\qquad T_z(1,-2,2)
  1. Remove the z-constant term: The term x squared y has zero z-partial derivative.
    ∂∂z(x2y)=0\frac{\partial}{\partial z}(x^2y)=0
  2. Differentiate the z term: Keep y and apply the power rule to z cubed.
    Tz(x,y,z)=3yz2T_z(x,y,z)=3yz^2
  3. Evaluate: Insert y = -2 and z = 2. The x-coordinate does not enter this derivative.
    Tz(1,−2,2)=3(−2)(22)=−24T_z(1,-2,2)=3(-2)(2^2)=-24

Answer

Tz(1,−2,2)=−24T_z(1,-2,2)=-24

Check: At x = 1 and y = -2, T becomes -2 - 2z cubed. Its ordinary z-derivative is -6z squared, which equals -24 at z = 2.

A mixed partial derivative

Differentiate once in x, then differentiate that result in y. Keeping the order visible prevents a variable-role mistake.

p(x,y)=x2y3+sin⁡(xy),pxyp(x,y)=x^2y^3+\sin(xy),\qquad p_{xy}
  1. Differentiate in x: Hold y fixed and apply the power and chain rules.
    px=2xy3+ycos⁡(xy)p_x=2xy^3+y\cos(xy)
  2. Differentiate the first term in y: Now x is fixed while y changes.
    ∂∂y(2xy3)=6xy2\frac{\partial}{\partial y}(2xy^3)=6xy^2
  3. Differentiate the product y cos(xy): Use the product rule and the chain rule.
    ∂∂y[ycos⁡(xy)]=cos⁡(xy)−xysin⁡(xy)\frac{\partial}{\partial y}[y\cos(xy)]=\cos(xy)-xy\sin(xy)

Answer

pxy=6xy2+cos⁡(xy)−xysin⁡(xy)p_{xy}=6xy^2+\cos(xy)-xy\sin(xy)

Check: Computing p sub y first and then differentiating in x produces the same expression, consistent with equality of mixed partials for this smooth function.

Try it yourself

Practice the decision before checking the answer

  1. Find f sub x.

    f(x,y)=4x3y−y2f(x,y)=4x^3y-y^2
    Show answer and explanation
    fx=12x2yf_x=12x^2y

    Hold y fixed and apply the x power rule.

  2. Find g sub y.

    g(x,y)=cos⁡(xy)g(x,y)=\cos(xy)
    Show answer and explanation
    gy=−xsin⁡(xy)g_y=-x\sin(xy)

    The inner derivative of xy with respect to y is x.

  3. Find h sub z.

    h(x,y,z)=zx+yh(x,y,z)=\frac{z}{x+y}
    Show answer and explanation
    hz=1x+yh_z=\frac1{x+y}

    The result requires x + y not equal to zero.

  4. Find q sub y at (3,0).

    q(x,y)=x2eyq(x,y)=x^2e^y
    Show answer and explanation
    qy(3,0)=9q_y(3,0)=9

    First q sub y equals x squared e to the y, then substitute.

Common mistakes

Notice the decision before the algebra

Changing every coordinate

Differentiating y while computing f sub x.

Fix: Write a short note such as y constant before applying any rule.

Holding the wrong variable fixed

Treating x as constant in an x-partial derivative.

Fix: Match the denominator of the partial notation to the one variable allowed to change.

Evaluating before differentiating

Substituting the point first and turning the function into a constant.

Fix: Find the symbolic partial derivative, keep its domain, and substitute coordinates last.

Confusing partial and implicit derivatives

Adding dy over dx to a function whose x and y inputs are independent.

Fix: Use partial notation for a multivariable function and implicit notation only when an equation makes y depend on x.

Assuming partials prove differentiability

Concluding that a function has a tangent plane because two partial derivatives exist at one point.

Fix: Check an appropriate differentiability condition, such as continuity of the partial derivatives in a neighborhood, before making that conclusion.