Topic 10 / 17intermediate

Partial Derivatives & Chain Rule

Compute partial derivatives, mixed partials, and use the chain rule to handle compositions involving several variables.

A partial derivative measures how $f$ changes when one variable changes and the others are held fixed. The geometric meaning is the slope of the surface in a chosen direction. Mixed partials and the chain rule extend single-variable calculus to functions of several variables.

Computing partial derivatives

f_{x} (x, y) = \frac{\partial f}{\partial x}

: differentiate with respect to

x

, treating

y

as a constant. Likewise

f_{y}

. The same rules from one-variable calculus apply (power, product, quotient, chain).

Key takeaways

Hold all other variables constant when differentiating.
Notation: $f_{x} = \partial_{x} f = \frac{\partial f}{\partial x}$ .

Limit definition (when forced)

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

. Use this only when explicitly required (e.g., proofs, non-elementary functions).

Higher-order and mixed partials

f_{xx} = \partial^{2} f / \partial x^{2}

f_{y y}

f_{x y} = (f_{x})_{y}

f_{y x} = (f_{y})_{x}

. **Clairaut/Schwarz Theorem**: if

f_{x y}

and

f_{y x}

are continuous,

f_{x y} = f_{y x}

. Mixed partials usually agree.

Chain rule — single intermediate variable

z = f (x, y)

and

x = x (t), y = y (t)

, then

\frac{d z}{d t} = \frac{\partial f}{\partial x} \frac{d x}{d t} + \frac{\partial f}{\partial y} \frac{d y}{d t}

Chain rule — multiple parameters

z = f (x, y)

with

x = x (s, t), y = y (s, t)

, then

\frac{\partial z}{\partial s} = f_{x} \frac{\partial x}{\partial s} + f_{y} \frac{\partial y}{\partial s}, \frac{\partial z}{\partial t} = f_{x} \frac{\partial x}{\partial t} + f_{y} \frac{\partial y}{\partial t} .

A **tree diagram** of dependencies prevents missing terms.

Strategy tips

Always draw a dependency tree when chain-ruling beyond two levels.
Each path from $z$ to a base variable contributes a product of partials.

Implicit differentiation

F (x, y, z) = 0

defines

z

implicitly as a function of

x, y

, then

\frac{\partial z}{\partial x} = - \frac{F _{x}}{F _{z}}

and similarly for

\partial z / \partial y

Worked examples

Example 1

Find

f_{x}

and

f_{y}

for

f (x, y) = x^{2} ln (x^{2} + y^{2})

1
$f_{x}$ : product rule, treating $y$ constant.
$f_{x} = 2 x ln (x^{2} + y^{2}) + x^{2} \cdot \frac{2 x}{x ^{2} + y ^{2}} = 2 x ln (x^{2} + y^{2}) + \frac{2 x ^{3}}{x ^{2} + y ^{2}}$
2
$f_{y}$ : $x^{2}$ is constant; chain through the log.
$f_{y} = x^{2} \cdot \frac{2 y}{x ^{2} + y ^{2}} = \frac{2 x ^{2} y}{x ^{2} + y ^{2}}$

Answer

f_{x} = 2 x ln (x^{2} + y^{2}) + \frac{2 x ^{3}}{x ^{2} + y ^{2}}

f_{y} = \frac{2 x ^{2} y}{x ^{2} + y ^{2}}

Example 2

Compute all second partials of

v = r cos (s + 2 t)

, where

r, s, t

are independent.

1
First partials.
$v_{r} = cos (s + 2 t), v_{s} = - r sin (s + 2 t), v_{t} = - 2 r sin (s + 2 t)$
2
Pure second partials.
$v_{r r} = 0, v_{ss} = - r cos (s + 2 t), v_{tt} = - 4 r cos (s + 2 t)$
3
Mixed (verify Clairaut).
$v_{r s} = - sin (s + 2 t) = v_{sr}, v_{r t} = - 2 sin (s + 2 t) = v_{t r}, v_{s t} = - 2 r cos (s + 2 t) = v_{t s}$

Answer

v_{r r} = 0

v_{ss} = - r cos (s + 2 t)

v_{tt} = - 4 r cos (s + 2 t)

, with mixed partials matching by Clairaut.

Example 3

u = x^{2} y^{3} + z^{4}

where

x = p + 3 p^{2}

y = e^{p}

z = cos p

, find

d u / d p

p = 0

1
Partial derivatives of $u$ .
$u_{x} = 2 x y^{3}, u_{y} = 3 x^{2} y^{2}, u_{z} = 4 z^{3}$
2
Derivatives of $x, y, z$ w.r.t. $p$ .
$x^{'} (p) = 1 + 6 p, y^{'} (p) = e^{p}, z^{'} (p) = - sin p$
3
Evaluate at $p = 0$ (so $x = 0, y = 1, z = 1$ ).
$u_{x} = 0, u_{y} = 0, u_{z} = 4; x^{'} (0) = 1, y^{'} (0) = 1, z^{'} (0) = 0$
4
Chain rule.
$\frac{d u}{d p}_{p = 0} = 0 \cdot 1 + 0 \cdot 1 + 4 \cdot 0 = 0$

Answer

d u / d p ∣_{p = 0} = 0

Interactive visualizations

Loading visualization…

x₀1.00

y₀1.00

f(x, y) = x² − y²/2

f_{x} (1.00, 1.00) = 2.000 (slope of blue x-trace at red dot)

f_{y} (1.00, 1.00) = - 1.000 (slope of green y-trace at red dot)

Drag the red point: blue line is the slope

f_{x}

(along

x

), orange is

f_{y}

(along

y

). The green and blue traces show

f

as a function of one variable with the other fixed.

Formulas in this topic

Partial derivative (limit definition)

f_{x} = h \to 0 lim \frac{f ( x + h , y ) - f ( x , y )}{h}

Definition of partial derivative.

Chain rule (one parameter)

\frac{d z}{d t} = f_{x} \frac{d x}{d t} + f_{y} \frac{d y}{d t}

z = f(x,y), x and y depend on t.

Chain rule (two parameters)

\frac{\partial z}{\partial s} = f_{x} x_{s} + f_{y} y_{s}

z = f(x,y), x and y depend on s, t.

Implicit differentiation

\frac{\partial z}{\partial x} = - \frac{F _{x}}{F _{z}}

When F(x,y,z) = 0 defines z implicitly.

Equality of mixed partials

f_{x y} = f_{y x}

If both are continuous (Clairaut).

Computing partial derivatives

Limit definition (when forced)

Higher-order and mixed partials

Chain rule — single intermediate variable

Chain rule — multiple parameters

Implicit differentiation

Worked examples

Interactive visualizations

Formulas in this topic

Related topics