Chapter 15

Hard

Partial Derivatives

00 · Symbol Glossary

$f(x,y)$f of x, y — function of two variables

A rule assigning one output number to each pair of inputs $(x,y)$ . Extends Chapter 01's single-variable $f(x)$ to two independent inputs; its graph is a surface in three dimensions rather than a curve in two.

$\dfrac{\partial f}{\partial x}$partial f partial x — partial derivative

The rate of change of $f$ with respect to $x$ alone, holding every other variable fixed. The rounded $\partial$ (distinct from Chapter 02's straight $d$ ) signals that other variables are present but frozen during this differentiation.

$f_x,\ f_y$f sub x, f sub y — subscript notation for partials

Compact alternatives to $\dfrac{\partial f}{\partial x}$ and $\dfrac{\partial f}{\partial y}$ , used when writing many partial derivatives becomes cumbersome.

$f_{xx},\ f_{xy}$second-order partial derivatives

$f_{xx}$ differentiates with respect to $x$ twice in a row. $f_{xy}$ (a mixed partial) differentiates first with respect to $y$ , then with respect to $x$ — order matters in the notation, though Section 03 shows it usually doesn't matter in the result.

$\nabla f$del f, or grad f — gradient

The directed quantity (Chapter 14) $\nabla f = \langle f_x, f_y\rangle$ , built from all of $f$ 's first partial derivatives. Points in the direction $f$ increases fastest at a given point.

$D_{\hat u}f$directional derivative of f in direction u-hat

The rate of change of $f$ in an arbitrary direction $\hat u$ (a unit vector, Chapter 14), not just along the $x$ - or $y$ -axis. Generalizes $f_x$ and $f_y$ , which are the special cases $\hat u = \hat\imath$ and $\hat u=\hat\jmath$ .

01 · Functions of Several Variables and Partial Derivatives

A function like $f(x,y) = x^2+y^2$ assigns a number to every point in the plane, not just to every point on a line. Its graph is a surface hovering over the $xy$ -plane. Differentiating such a function requires picking one direction to move in at a time.

Definition — Partial Derivative

\frac{\partial f}{\partial x} = \lim_{h\to0}\frac{f(x+h,y)-f(x,y)}{h} \qquad \frac{\partial f}{\partial y} = \lim_{h\to0}\frac{f(x,y+h)-f(x,y)}{h}

Each is Chapter 02's difference-quotient definition, applied to only one variable at a time, with the other variable frozen as if it were a constant.

Step-by-step — Find $f_x$ and $f_y$ for $f(x,y) = x^3y^2 + 4xy$

Compute $f_x$ — differentiate with respect to $x$ , treating $y$ as a constant: the term

x^3y^2

becomes

3x^2y^2

(power rule on

x^3

, Chapter 02;

y^2

is just a constant multiplier). The term

4xy

becomes

4y

(

x

's coefficient is

4y

, a constant with respect to

x

Assemble $f_x$ :

f_x = 3x^2y^2+4y

Compute $f_y$ — differentiate with respect to $y$ , treating $x$ as a constant: the term

x^3y^2

becomes

2x^3y

(power rule on

y^2

;

x^3

is a constant multiplier). The term

4xy

becomes

4x

Assemble $f_y$ :

f_y = 2x^3y+4x

Evaluate both at $(x,y)=(1,2)$ :

f_x(1,2) = 3(1)(4)+4(2) = 12+8=20

f_y(1,2) = 2(1)(2)+4(1)=4+4=8

Geometric meaning — slicing the surface

Fixing $y=y_0$ and letting $x$ vary slices the surface $z=f(x,y)$ with a vertical plane, producing an ordinary curve $z=f(x,y_0)$ in that slice. $f_x(x_0,y_0)$ is exactly the slope of that sliced curve at $x=x_0$ — an ordinary single-variable derivative (Chapter 02), just computed on the curve left behind by one particular slice. $f_y$ is the same idea, slicing the other way.

❌ Forgetting to hold the other variable fixed

For $f(x,y)=x^2y$ , computing $f_x$ by treating $y$ as if it were also varying with $x$ — e.g. mistakenly applying the product rule to $x^2$ and $y$ as if $y=y(x)$ , giving $2xy + x^2\dfrac{dy}{dx}$ .

Why it breaks: the entire point of a partial derivative is that $y$ is treated as a fixed constant while differentiating with respect to $x$ — there is no $\dfrac{dy}{dx}$ term at all, because $y$ is not being allowed to change during this computation.

Consequence: the correct partial derivative is simply $f_x = 2xy$ (power rule on $x^2$ , with $y$ as a constant multiplier) — no product rule needed, since only one factor ( $x^2$ ) actually depends on the variable being differentiated.

02 · Higher-Order Partial Derivatives

Definition — Second-Order Partial Derivatives

f_{xx} = \frac{\partial}{\partial x}\left(\frac{\partial f}{\partial x}\right) \qquad f_{yy} = \frac{\partial}{\partial y}\left(\frac{\partial f}{\partial y}\right) \qquad f_{xy} = \frac{\partial}{\partial x}\left(\frac{\partial f}{\partial y}\right) \qquad f_{yx} = \frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x}\right)

Reading order for mixed partials: $f_{xy}$ means differentiate with respect to $y$ first, then $x$ — the rightmost subscript is applied first, since it's the inner operation.

Step-by-step — Find all second-order partials of $f(x,y)=x^3y^2$

First partials (Section 01's method):

f_x = 3x^2y^2

;

f_y = 2x^3y

$f_{xx}$ — differentiate $f_x$ with respect to $x$ again:

6xy^2

(power rule,

y^2

constant).

$f_{yy}$ — differentiate $f_y$ with respect to $y$ again:

2x^3

(power rule,

x^3

constant).

$f_{xy}$ — differentiate $f_y=2x^3y$ with respect to $x$ :

6x^2y

$f_{yx}$ — differentiate $f_x=3x^2y^2$ with respect to $y$ :

6x^2y

Compare steps 4 and 5:

f_{xy}=f_{yx}=6x^2y

— the two mixed partials agree, despite being computed in opposite orders.

Clairaut's Theorem

Step 6 is not a coincidence: if $f_{xy}$ and $f_{yx}$ are both continuous near a point, they are guaranteed equal there. This holds for essentially every function built from polynomials, and from $e^x$ , $\sin x$ , $\cos x$ , $\ln x$ (Chapter 02) combined by sums, products, and composition — order of mixed partial differentiation almost never matters in practice, though the continuity hypothesis is technically required.

03 · The Gradient

Bundling both partial derivatives into a single directed quantity (Chapter 14) produces the gradient — the tool that identifies the single best direction to move in.

Definition — Gradient

\nabla f(x,y) = \langle f_x(x,y),\ f_y(x,y)\rangle

At each point, $\nabla f$ is a directed quantity (Chapter 14) pointing in the direction $f$ increases most rapidly, with magnitude $\lVert\nabla f\rVert$ equal to that maximum rate of increase.

Step-by-step — Find $\nabla f$ at $(1,2)$ for $f(x,y) = x^2+3xy$

Compute $f_x$ : treating

y

as constant,

f_x = 2x+3y

Compute $f_y$ : treating

x

as constant,

f_y = 3x

Assemble the gradient as a directed quantity (Chapter 14 component form):

\nabla f = \langle2x+3y,\ 3x\rangle

Evaluate at $(1,2)$ :

\nabla f(1,2) = \langle2(1)+3(2),\ 3(1)\rangle = \langle8,3\rangle

04 · The Directional Derivative

$f_x$ and $f_y$ measure the rate of change only along the two axis directions. The directional derivative generalizes this to any direction at all, using the dot product (Chapter 14) to combine the gradient with a chosen direction.

Definition — Directional Derivative

For a unit vector $\hat u$ (Chapter 14: $\lVert\hat u\rVert=1$ ):

D_{\hat u}f = \nabla f \cdot \hat u

$D_{\hat u}f$ is the rate of change of $f$ as position moves in the direction $\hat u$ .

Step-by-step — Find the directional derivative of $f(x,y)=x^2+3xy$ at $(1,2)$ in the direction of $\vec v=\langle3,4\rangle$

**Recall

\nabla f(1,2) = \langle8,3\rangle

from Section 03.

Normalize $\vec v$ into a unit vector (Chapter 14, Section 02's magnitude formula):

\lVert\vec v\rVert = \sqrt{3^2+4^2}=\sqrt{25}=5

;

\hat u = \dfrac{\vec v}{\lVert\vec v\rVert} = \left\langle\dfrac35,\dfrac45\right\rangle

Apply the dot product formula (Chapter 14, Section 03):

D_{\hat u}f = \nabla f\cdot\hat u = \langle8,3\rangle\cdot\left\langle\dfrac35,\dfrac45\right\rangle = 8\left(\dfrac35\right)+3\left(\dfrac45\right) = \dfrac{24}{5}+\dfrac{12}{5}=\dfrac{36}{5}

Conclude: moving from

(1,2)

in the direction of

\langle3,4\rangle

f

increases at rate

\dfrac{36}{5}=7.2

per unit distance traveled.

❌ Using a non-unit vector directly in the directional derivative formula

On the same problem, skipping step 2 and computing $\nabla f\cdot\vec v = \langle8,3\rangle\cdot\langle3,4\rangle = 24+12=36$ directly.

Why it breaks: the directional derivative's definition (Section 04) explicitly requires $\hat u$ to be a unit vector — a direction of length exactly $1$ , carrying no scaling of its own. Using $\vec v=\langle3,4\rangle$ directly (with $\lVert\vec v\rVert=5\neq1$ ) mixes the direction together with an extra factor of $5$ from its own length.

Consequence: $36 \neq \dfrac{36}{5}$ (the correctly normalized answer from step 3) — the un-normalized result is exactly $5\times$ too large, since it silently multiplied the true directional derivative by $\lVert\vec v\rVert=5$ . Always normalize the direction vector to length $1$ before applying the dot product with $\nabla f$ .

05 · The Tangent Plane and the Multivariable Chain Rule

Definition — Tangent Plane

At a point $(x_0,y_0)$ where $f$ is differentiable, the plane tangent to the surface $z=f(x,y)$ is

z = f(x_0,y_0) + f_x(x_0,y_0)(x-x_0) + f_y(x_0,y_0)(y-y_0)

directly extending Chapter 02, Section 02's tangent-line formula ( $y=f(a)+f'(a)(x-a)$ ) with one extra term for the second variable.

Definition — Multivariable Chain Rule (along a parametrized path)

If $x=x(t)$ and $y=y(t)$ (Chapter 13's parametric functions), and $z=f(x,y)$ , then

\frac{dz}{dt} = f_x\cdot\frac{dx}{dt} + f_y\cdot\frac{dy}{dt}

Each term applies Chapter 03's single-variable chain rule along one "slice" direction, and the two contributions add — the total rate of change from both $x$ and $y$ moving simultaneously.

Step-by-step — Find $\dfrac{dz}{dt}$ for $z=x^2y$, $x(t)=t$, $y(t)=t^2$, at $t=2$

Compute the partials of $z$ :

f_x = 2xy

;

f_y = x^2

Compute $\dfrac{dx}{dt}$ and $\dfrac{dy}{dt}$ (Chapter 02):

\dfrac{dx}{dt}=1

;

\dfrac{dy}{dt}=2t

At $t=2$ : find $x,y$ :

x(2)=2

y(2)=4

Evaluate the partials at $(x,y)=(2,4)$ :

f_x(2,4)=2(2)(4)=16

;

f_y(2,4)=2^2=4

Evaluate $\dfrac{dx}{dt}$ , $\dfrac{dy}{dt}$ at $t=2$ :

\dfrac{dx}{dt}=1

;

\dfrac{dy}{dt}=2(2)=4

Apply the chain rule formula:

\dfrac{dz}{dt} = f_x\cdot\dfrac{dx}{dt}+f_y\cdot\dfrac{dy}{dt} = 16(1)+4(4) = 16+16=32

06 · Exercises

EXERCISE 15.1

Differentiate with respect to one variable while treating the other as a fixed constant, exactly as in Section 01.

$f(x,y)=4x^2y^3-2y$ . $f_x = 8xy^3$ (power rule, $y^3$ constant). $f_y = 12x^2y^2-2$ (power rule on $y^3$ and $y$ ; $4x^2$ constant).

Find $f_x$ and $f_y$ for $f(x,y)=4x^2y^3-2y$ .

EXERCISE 15.2

Compute both first partials first, then differentiate each of those again with respect to the appropriate variable.

$f(x,y)=x^4y^2$ . $f_x=4x^3y^2$ , $f_y=2x^4y$ . $f_{xx}=12x^2y^2$ . $f_{yy}=2x^4$ . $f_{xy}=\frac{\partial}{\partial x}(2x^4y)=8x^3y$ . $f_{yx}=\frac{\partial}{\partial y}(4x^3y^2)=8x^3y$ . The two mixed partials match, as Clairaut's theorem (Section 02) guarantees.

Find all four second-order partial derivatives of $f(x,y)=x^4y^2$ , and confirm $f_{xy}=f_{yx}$ .

EXERCISE 15.3

Compute both first partials, evaluate them at the given point, and assemble the result as a directed quantity in component form.

$f(x,y)=xy^2-x^2$ at $(2,1)$ . $f_x=y^2-2x$ , $f_y=2xy$ . At $(2,1)$ : $f_x=1-4=-3$ ; $f_y=4$ . $\nabla f(2,1) = \langle-3,4\rangle$ .

Find $\nabla f$ at $(2,1)$ for $f(x,y)=xy^2-x^2$ .

EXERCISE 15.4

Normalize the direction vector to a unit vector before taking the dot product with the gradient.

Using $\nabla f(2,1)=\langle-3,4\rangle$ from Exercise 15.3 and direction $\vec v=\langle1,0\rangle$ : $\lVert\vec v\rVert=1$ already (it's already a unit vector). $D_{\hat u}f = \langle-3,4\rangle\cdot\langle1,0\rangle = -3(1)+4(0)=-3$ .

Using the gradient from Exercise 15.3, find the directional derivative of $f$ at $(2,1)$ in the direction $\vec v=\langle1,0\rangle$ .

EXERCISE 15.5

Compute the two partial derivatives, evaluate at the given point, then plug into the tangent plane formula.

$f(x,y)=x^2+y^2$ at $(1,1)$ : $f(1,1)=2$ . $f_x=2x$ , $f_y=2y$ ; at $(1,1)$ : $f_x=2$ , $f_y=2$ . Tangent plane: $z = 2+2(x-1)+2(y-1) = 2+2x-2+2y-2 = 2x+2y-2$ .

Find the equation of the tangent plane to $z=x^2+y^2$ at $(1,1)$ .

EXERCISE 15.6

Compute the partials of the value function and the parametric derivatives of each input, then combine using the multivariable chain rule.

A portfolio's value is $V(x,y)=50x+30y$ dollars where $x,y$ are two asset prices, both changing with time: $x(t)=10+2t$ , $y(t)=20-t$ . $V_x=50$ , $V_y=30$ . $\dfrac{dx}{dt}=2$ , $\dfrac{dy}{dt}=-1$ . $\dfrac{dV}{dt} = 50(2)+30(-1) = 100-30=70$ . Portfolio value is increasing at $70$ dollars per unit time.

A portfolio's value is $V(x,y)=50x+30y$ dollars, where $x(t)=10+2t$ and $y(t)=20-t$ are two asset prices changing with time $t$ . Find $\dfrac{dV}{dt}$ using the multivariable chain rule.

07 · Chapter Summary

Concept	Formula / Rule
Partial derivative	$f_x = \lim_{h\to0}\frac{f(x+h,y)-f(x,y)}{h}$ ; other variable held fixed
Mixed partials	$f_{xy}=f_{yx}$ when both are continuous (Clairaut's Theorem)
Gradient	$\nabla f = \langle f_x, f_y\rangle$ ; points toward steepest increase
Directional derivative	$D_{\hat u}f = \nabla f\cdot\hat u$ , $\hat u$ a unit vector
Tangent plane	$z=f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)$
Multivariable chain rule	$\dfrac{dz}{dt} = f_x\dfrac{dx}{dt}+f_y\dfrac{dy}{dt}$

Next: Chapter 16 — Multiple Integrals, which extends the definite integral from area under a curve to volume under a surface.