Calculus Volume 3 · Chapter 4 · Section 4.3

Partial Derivatives

A surface has a slope in every direction. Freeze all variables but one, and the ordinary derivative gives the slope along that one.

Idea 1 · freeze one variable $f_x$ : hold $y$ fixed, differentiate in $x$ .Every rule from one variable still works. Idea 2 · slice the surface $f_x(a,b)$ is the slope of a curve.The trace in the plane

y=b

. Idea 3 · order rarely matters $f_{xy}=f_{yx}$ when both are continuous.

Colour key surface along $x$ : $f_x$ along $y$ : $f_y$ tangent, secant key idea · 3D figures turn when dragged · page links open the textbook

L1One point, two slopes
L1Freeze one variable
L1Slice the surface
L2Definitions
L2Examples 4.14–4.20
L2Second partials and Clairaut
L2Partial differential equations
L2Lab 1 · slice and slope
L2Lab 2 · a secant from a contour map
L2Lab 3 · two orders, one answer
L2Lab 4 · a cooling wire
L2Common mistakes
L2Practice and answers
L3Why freezing works
L3When the order does matter
L3Why sharp heat dies first
L3Two slopes make a plane

Level 1 · see itSlopes of a surfacebefore any formula

One point, two slopes

A hill $z=\sqrt{9-x^2-y^2}$ , drawn as a contour map (Figure 4.22). You stand at $(1,2)$ , height 2.

Step east: the ground drops $\tfrac12$ per step. Step north: it drops 1.

A surface has no single slope. Name a direction first. The two easiest directions give the two partial derivatives.

Level curves

z=0,1,2,2.5

. east: $-\tfrac12$ , north: $-1$ .

Freeze one variable

$\partial f/\partial x$ , or $f_x$ : treat $y$ as a constant and differentiate in $x$ (p331).
$\partial f/\partial y$ , or $f_y$ : treat $x$ as a constant. The rounded $\partial$ warns that other variables are standing still.
Example 4.15a: $f=x^2-3xy+2y^2-4x+5y-12$ gives $f_x=2x-3y-4$ and $f_y=-3x+4y+5$ (p332).

In $f_x$ , the terms $2y^2$ , $5y$ and $-12$ are constants, so they vanish; $-3xy$ is $-3y$ times $x$ .

Slice the surface

Cut the hill with the plane $y=2$ . The cut is a curve, $z=\sqrt{5-x^2}$ ; its slope at $x=1$ is $f_x(1,2)=-\tfrac12$ .

Cut with $x=1$ instead: $z=\sqrt{8-y^2}$ , slope $f_y(1,2)=-1$ (p333).

A partial derivative is the slope of a slice.

Plane $y=2$ : slope

-\tfrac12

. Plane $x=1$ : slope

-1

. Red: the tangent lines through

(1,2,2)

. Drag to turn.

Level 2 · compute itLimits, rules, second partialsand four labs

Definitions

Partial in x · (4.12) · p330

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

Only $x$ moves.

Partial in y · (4.13)

f_y(x,y)=\lim_{k\to0}\frac{f(x,y+k)-f(x,y)}{k}

Only $y$ moves.

Three variables · (4.14)–(4.16) · p335

$f_x$ , $f_y$ , $f_z$ : move one of $x,y,z$ , freeze the other two.

From a contour map · p333

Rise between two level curves over the run between them: a secant slope.

The limit is the one-variable derivative of $g(x)=f(x,b)$ ; that is why every familiar rule applies.

Examples 4.14–4.20

Example	given	result
4.14 · p330	$x^2-3xy+2y^2-4x+5y-12$ , by limits	$f_x=2x-3y-4$ , $f_y=-3x+4y+5$
4.15 · p332	same $f$ ; $g=\sin(x^2y-2x+4)$	same answers; $g_x=(2xy-2)\cos(\cdot)$ , $g_y=x^2\cos(\cdot)$
4.16 · p333	$g=\sqrt{9-x^2-y^2}$ , $g_x$ at $(\sqrt5,0)$	contour estimate $\approx-1.688$ ; exact $-\sqrt5/2\approx-1.118$
4.17 · p335	$x^2-3xy+2y^2-4xz+5yz^2-12x+4y-3z$	$f_x=2x-3y-4z-12$ , $f_y=-3x+4y+5z^2+4$ , $f_z=-4x+10yz-3$
4.18 · p336	$\dfrac{x^2y-4xz+y^2}{x-3yz}$ ; $\sin(x^2y-z)+\cos(x^2-yz)$	quotient rule; chain rule, e.g. $g_z=-\cos(x^2y-z)+y\sin(x^2-yz)$
4.19 · p337	$xe^{-3y}+\sin(2x-5y)$	$f_{xx}=-4\sin(2x-5y)$ , $f_{xy}=f_{yx}=-3e^{-3y}+10\sin(2x-5y)$ , $f_{yy}=9xe^{-3y}-25\sin(2x-5y)$
4.20 · p339	$5\sin(3\pi x)\sin(4\pi y)\cos(10\pi t)$	$u_{tt}=-500\pi^2(\cdot)=4(u_{xx}+u_{yy})$

Second partials and Clairaut

Four second partials · p337

f_{xx},\quad f_{xy}=\frac{\partial}{\partial y}\Big[\frac{\partial f}{\partial x}\Big],\quad f_{yx},\quad f_{yy}

Subscripts read left to right: $f_{xy}$ is $x$ first, then $y$ .

Theorem 4.5 · Clairaut

If $f_{xy}$ and $f_{yx}$ are continuous on an open disk around $(a,b)$ , then

f_{xy}(a,b)=f_{yx}(a,b).

Mixed partials measure how the $x$ -slope changes as you step in $y$ — and that is the same as how the $y$ -slope changes as you step in $x$ .

Partial differential equations

Heat · (4.17) · p339

$u_t=c^2(u_{xx}+u_{yy})$

Wave · (4.18)

$u_{tt}=c^2(u_{xx}+u_{yy})$

Laplace · (4.19)

$u_{xx}+u_{yy}=0$

A partial differential equation links an unknown function of several variables to its partial derivatives. On a wire, $u_t=c^2u_{xx}$ has the solutions $u_m=e^{-\pi^2m^2c^2t}\sin(m\pi x)$ ((4.22)).

To check a solution, compute each partial and substitute; nothing more.

Lab 1 · slice and slope

surface

slope along

a1 b2

Try: the saddle at $a=0$ — along $x$ the slice is a valley, along $y$ a ridge, both flat at the centre. The slope depends on the point and on the direction; the 2D graph below is the slice laid flat.

Lab 2 · a secant from a contour map

case

step h0.59

Try: the next level curve (the book's estimate), then shrink $h$ . A contour map gives a secant; only the limit gives the tangent.

Lab 3 · two orders, one answer

function

square side h0.3

Drag the corner

(a,b)

. Corner signs:

+\,-\,-\,+

Try: the third function with the corner at the origin. x-first and y-first disagree there: the mixed partials exist but are not continuous, so Clairaut does not apply.

Lab 4 · a cooling wire

mode m

time t0 c²1

u=e^{-\pi^2m^2c^2t}\sin(m\pi x)

over

0\le x\le1

0\le t\le0.3

(Figure 4.24).

Try: mode 3 against mode 1. A wiggly profile has a large $u_{xx}$ , so $u_t=c^2u_{xx}$ makes it fade fast.

Common mistakes

Mistake	Result	Fix
Differentiating the frozen variable	$f_x$ of $2y^2$ written as $4y$	a term without $x$ is a constant: 0
Dropping the inner derivative	$g_x=\cos(x^2y-2x+4)$	chain rule: times $2xy-2$
$f_{xy}$ read as "y first"	order swapped when it matters	subscripts left to right; $\partial^2 f/\partial y\,\partial x$ right to left
Contour estimate taken as exact	$-1.688$ for $-1.118$	it is a secant; differentiate for the tangent
Clairaut used without continuity	$f_{xy}(0,0)=f_{yx}(0,0)$ "proved" for the counterexample	check the hypothesis (p338)
Quotient rule in the wrong order	sign error in Example 4.18a	$(\text{top}'\cdot\text{bottom}-\text{top}\cdot\text{bottom}')/\text{bottom}^2$

Practice and answers

Exercises from p344; the book's key: p862, p867–868.

Checkpoints 4.12–4.18

#	task	answer
4.12	$4x^2+2xy-y^2+3x-2y+5$ , by limits	$f_x=8x+2y+3$ , $f_y=2x-2y-2$
4.13	$\tan(x^3-3x^2y^2+2y^4)$	$f_x=(3x^2-6xy^2)\sec^2(\cdot)$ , $f_y=(-6x^2y+8y^3)\sec^2(\cdot)$
4.14	$x^2-y^2$ , $f_y$ at $(0,\sqrt2)$ from contours	$\approx-\sqrt3-\sqrt2\approx-3.146$ ; exact $-2\sqrt2\approx-2.828$
4.15	$2x^2-4x^2y+2y^2+5xz^2-6x+3z-8$	$f_x=4x-8xy+5z^2-6$ , $f_y=-4x^2+4y$ , $f_z=10xz+3$
4.16	$\sec(x^2y)-\tan(x^3yz^2)$	$f_z=-2x^3yz\sec^2(x^3yz^2)$ ; $f_x$ , $f_y$ in the key
4.17	$\sin(3x-2y)+\cos(x+4y)$	$f_{xx}=-9\sin(3x-2y)-\cos(x+4y)$ ; $f_{xy}=f_{yx}=6\sin(3x-2y)-4\cos(x+4y)$ ; $f_{yy}=-4\sin(3x-2y)-16\cos(x+4y)$
4.18	$2\sin\tfrac x3\sin\tfrac y4\,e^{-25t/16}$ solves $u_t=9(u_{xx}+u_{yy})$	$u_t=-\tfrac{25}{16}u$ , $9\big(-\tfrac19-\tfrac1{16}\big)u=-\tfrac{25}{16}u$ (not in the key)

Exercises

#	task	answer	key step
113	$z_y$ , $z=x^2-3xy+y^2$ , by limits	$-3x+2y$	$z(x,y+k)-z=-3xk+2yk+k^2$
115, 117	signs from the bowl's graph	$f_x(-1,1)<0$ ; $f_x(0,0)=0$	slope of the slice
119	$z_y$ , $z=\sin3x\cos3y$	$-3\sin(3x)\sin(3y)$	$\sin3x$ is a constant
121	$\ln(x^6+y^4)$	$\dfrac{6x^5}{x^6+y^4}$ , $\dfrac{4y^3}{x^6+y^4}$	chain rule
125	$\tan(2x-y)$	$2\sec^2(2x-y)$ , $-\sec^2(2x-y)$	inner derivatives 2, −1
127	$\arctan(y/x)$ at $(2,-2)$	$f_x=f_y=\tfrac14$	$f_x=-y/(x^2+y^2)$
129	$z_x$ of $e^{-x}\cos y$ at $(0,1)$	$-\cos1$	—
133	cylinder $V(r,h)$	$\pi r^2h$ ; $V_r=2\pi rh$ ; $V_h=\pi r^2$	circumference × height; base area
135	$f_{xy}$ , $z=\ln(x-y)$	$\dfrac1{(x-y)^2}$	$z_x=(x-y)^{-1}$
143	$f_{xyz}$ , $x^2y^3z-3xy^2z^3+5x^2z-y^3z$	$6xy^2-18yz^2$	one variable at a time
145	$f_x=f_y=0$ , $x^2+x-3xy+y^3-5$	$(\tfrac14,\tfrac12)$ , $(1,1)$	$x=y^2$ , then $2y^2-3y+1=0$
147	$f_x=f_y=0$ , $y^3-3yx^2-3y^2-3x^2+1$	$(0,0),(0,2),(\pm\sqrt3,-1)$	$f_x=-6x(y+1)$
151	$e^{-t}\cos(x/c)$ solves $z_t=c^2z_{xx}$	$c^2z_{xx}=-e^{-t}\cos(x/c)=z_t$	the key omits the minus sign
159	$F=x^2+y^2-2xy\cos\theta$ at $(2,3,\pi/6)$	$F_\theta=6$ , $F_x=4-3\sqrt3$	$F_\theta=2xy\sin\theta$
161	$200x^{0.7}y^{0.3}$ at $(500,1000)$	$f_x\approx172.36$ , $f_y\approx36.93$	$140(y/x)^{0.3}$ , $60(x/y)^{0.7}$

Level 3 · why it worksWhere the rules come fromshort arguments

Why freezing works

Fix $y=b$ and let $g(x)=f(x,b)$ . Then (p331)

g'(a)=\lim_{h\to0}\frac{g(a+h)-g(a)}{h}=\lim_{h\to0}\frac{f(a+h,b)-f(a,b)}{h}=f_x(a,b).

A partial derivative is an ordinary derivative of a one-variable function, so the power, product, quotient and chain rules come for free.

When the order does matter

Take $f=\dfrac{xy(x^2-y^2)}{x^2+y^2}$ , $f(0,0)=0$ . From the limits, $f_x(0,y)=-y$ and $f_y(x,0)=x$ , so

f_{xy}(0,0)=-1,\qquad f_{yx}(0,0)=1.

Away from the origin both mixed partials equal

\frac{(x^2-y^2)(x^4+10x^2y^2+y^4)}{(x^2+y^2)^3},

which has no limit at $(0,0)$ . Clairaut's continuity hypothesis is exactly what fails.

A smooth-looking surface whose mixed partials disagree at one point.

Why sharp heat dies first

For $u_m=e^{-\pi^2m^2c^2t}\sin(m\pi x)$ : $u_t=-\pi^2m^2c^2u$ and $u_{xx}=-\pi^2m^2u$ , so $u_t=c^2u_{xx}$ (Figure 4.24).

With $c^2=1$ , at $t=0.1$ mode 1 keeps $e^{-0.99}\approx0.37$ of its height; mode 3 keeps about $0.0001$ . Decay grows with $m^2$ : after a short time only the smoothest mode is left — the reason Kelvin needed only the first terms.

Two slopes make a plane

The two tangent lines at $(1,2,2)$ have directions $\langle1,0,f_x\rangle$ and $\langle0,1,f_y\rangle$ . They span a plane:

z=2-\tfrac12(x-1)-(y-2).

If the surface is smooth, this is its tangent plane — §4.4.

Equation	Name	Unknown
$u_t=c^2(u_{xx}+u_{yy})$	heat equation (2D)	$u(x,y,t)$
$u_{tt}=c^2(u_{xx}+u_{yy})$	wave equation (2D)	$u(x,y,t)$
$u_{xx}+u_{yy}=0$	Laplace's equation (2D)	$u(x,y)$

Objective	You can now
4.3.1 two variables	$f_x,f_y$ by the limit or by freezing; read them as slice slopes
4.3.2 more variables	$f_x,f_y,f_z$ : freeze all but one
4.3.3 higher order	$f_{xx},f_{xy},f_{yx},f_{yy}$ ; $f_{xy}=f_{yx}$ when continuous
4.3.4 PDEs	name heat, wave, Laplace; verify a solution by substitution

One point, two slopes

Freeze one variable

Slice the surface

Definitions

Examples 4.14–4.20

Second partials and Clairaut

Partial differential equations

Lab 1 · slice and slope

Lab 2 · a secant from a contour map

Lab 3 · two orders, one answer

Lab 4 · a cooling wire

Common mistakes

Practice and answers

Why freezing works

When the order does matter

Why sharp heat dies first

Two slopes make a plane

Partial Derivatives

A partial derivative is the ordinary derivative with the other variable held fixed

From the limit, every term without hhh cancels: fx=2x−3y−4f_x=2x-3y-4fx​=2x−3y−4

To find ∂f/∂x\partial f/\partial x∂f/∂x, treat every other variable as a constant

Chain rule: the inner derivative is taken in the moving variable only

fx(a,b)f_x(a,b)fx​(a,b) is the slope of the curve cut from the surface by the plane y=by=by=b

Secant slopes along the slice close in on fxf_xfx​: −1, −0.68, −0.53→−12-1,\ -0.68,\ -0.53\to-\tfrac12−1, −0.68, −0.53→−21​

fxf_xfx​ and fyf_yfy​ are the slopes of the tangent lines parallel to the xxx- and yyy-axes

A contour map estimates gxg_xgx​ as rise over run between level curves

For fyf_yfy​, run along yyy: fy(0,2)≈−3.146f_y(0,\sqrt2)\approx-3.146fy​(0,2​)≈−3.146, exact −2.828-2.828−2.828

With three variables, freeze two and differentiate the third

Quotient and chain rules work unchanged; frozen variables ride along as constants

Differentiate a partial again: four second-order partials, two of them mixed

Both orders give the same mixed partial: fxy=fyx=−3e−3y+10sin⁡(2x−5y)f_{xy}=f_{yx}=-3e^{-3y}+10\sin(2x-5y)fxy​=fyx​=−3e−3y+10sin(2x−5y)

If the mixed partials are continuous, the order of differentiation does not matter

One corner sum equals h2fxyh^2f_{xy}h2fxy​ at one point and h2fyxh^2f_{yx}h2fyx​ at another

Without continuity the mixed partials can differ: fxy(0,0)=−1f_{xy}(0,0)=-1fxy​(0,0)=−1, fyx(0,0)=1f_{yx}(0,0)=1fyx​(0,0)=1

A partial differential equation ties an unknown function of several variables to its partial derivatives

The wave equation checks out: both sides equal −500π2S-500\pi^2S−500π2S

Every partial is a multiple of uuu, and the multiples match: ut=9(uxx+uyy)u_t=9(u_{xx}+u_{yy})ut​=9(uxx​+uyy​)

In one dimension, ut=c2uxxu_t=c^2u_{xx}ut​=c2uxx​: the wire's hot middle cools fastest

Higher modes decay like e−π2m2te^{-\pi^2m^2t}e−π2m2t: sharp heat dies first

One variable moves, the rest are constants — and every Calc I rule applies

From the limit, every term without $h$ cancels: $f_x=2x-3y-4$

To find $\partial f/\partial x$ , treat every other variable as a constant

$f_x(a,b)$ is the slope of the curve cut from the surface by the plane $y=b$

Secant slopes along the slice close in on $f_x$ : $-1,\ -0.68,\ -0.53\to-\tfrac12$

$f_x$ and $f_y$ are the slopes of the tangent lines parallel to the $x$ - and $y$ -axes

A contour map estimates $g_x$ as rise over run between level curves

For $f_y$ , run along $y$ : $f_y(0,\sqrt2)\approx-3.146$ , exact $-2.828$

Both orders give the same mixed partial: $f_{xy}=f_{yx}=-3e^{-3y}+10\sin(2x-5y)$

One corner sum equals $h^2f_{xy}$ at one point and $h^2f_{yx}$ at another

Without continuity the mixed partials can differ: $f_{xy}(0,0)=-1$ , $f_{yx}(0,0)=1$

The wave equation checks out: both sides equal $-500\pi^2S$

Every partial is a multiple of $u$ , and the multiples match: $u_t=9(u_{xx}+u_{yy})$

In one dimension, $u_t=c^2u_{xx}$ : the wire's hot middle cools fastest

Higher modes decay like $e^{-\pi^2m^2t}$ : sharp heat dies first