4.1.2 Partial derivatives

The slope of a slice

Lesson 4.1.1 cut the bowl $f(x, y) = x^2 + 2y^2$ into slices. Fix $y = 1$ and the slice is $x^2 + 2$ , a function of $x$ alone, so it has a slope. That slope is a partial derivative: the rate at which $f$ changes as $x$ moves, with $y$ held fixed.

Take the point $(1, 1)$ , where $f = 3$ . Move $x$ by a gap $h$ and keep $y$ at 1. Then do the same in the other direction: keep $x$ at 1 and move $y$ .

gap $h$	$\pm 0.1$	$\pm 0.01$	$\pm 0.001$
chord slope, $x$ moving	2.1, 1.9	2.01, 1.99	2.001, 1.999
chord slope, $y$ moving	4.2, 3.8	4.02, 3.98	4.002, 3.998

From both sides the first row settles on 2 and the second on 4. So at $(1, 1)$ the surface rises twice as fast in the $y$ direction as in the $x$ direction.

The algebra agrees. With $y = 1$ fixed, the chord slope is $\frac{(1 + h)^2 + 2 - 3}{h} = 2 + h$ , which tends to 2. With $x = 1$ fixed, it is $\frac{1 + 2(1 + h)^2 - 3}{h} = 4 + 2h$ , which tends to 4.

Rule: Partial derivatives

\begin{gathered} \frac{\partial f}{\partial x}(a, b) = \lim_{h \to 0} \frac{f(a + h, b) - f(a, b)}{h} \\[6pt] \frac{\partial f}{\partial y}(a, b) = \lim_{h \to 0} \frac{f(a, b + h) - f(a, b)}{h} \end{gathered}

Lesson 2.2.2's derivative, taken along one input while the other stays fixed. $\frac{\partial f}{\partial x}$ is the slope of the slice $y = b$ at $x = a$ ; $\frac{\partial f}{\partial y}$ is the slope of the slice $x = a$ at $y = b$ .

The two slices of $x^2 + 2y^2$ through $(1, 1)$ , with their tangents at the point. The slope of each tangent is a partial derivative: 2 in the $x$ direction, 4 in the $y$ direction.

Try it on the explorer. The point starts at (2, 1) on the bowl $x^2 + 2y^2$ , where the slice along $x$ has slope 4. Slide the point along the dashed line and watch the slope change; then slice along $y$ instead, and try the other functions.

Explore

Hold y fixed and f becomes a function of x alone: a slice through the surface, graphed below. Its slope at the point is the partial derivative ∂f/∂x. Move the point, or slice along y instead to read ∂f/∂y.

The slice y = 1: f(x, 1) against x, with its tangent at the point

Point x = 2

Point y = 1

Slice along y instead (hold x fixed)

Point: (2, 1), f = 6
Partial derivatives: ∂f/∂x = 4 (the slope of the slice along x), ∂f/∂y = 4 (along y)

Notation

The curly $\partial$ is read "partial", so $\frac{\partial f}{\partial x}$ is "partial f by partial x". It replaces the $d$ of $\frac{df}{dx}$ to warn that $f$ has other inputs, held fixed. A shorter name is $f_x$ , with the input that moves written below.

Both name a new function of $x$ and $y$ . For the bowl, $f_x = 2x$ and $f_y = 4y$ at every point. At one point, write $f_x(1, 1) = 2$ , giving both coordinates.

(1, 1)

def f(x, y): return x**2 + 2 * y**2 for h in [0.1, 0.01, 0.001]: in_x = (f(1 + h, 1) - f(1, 1)) / h, (f(1 - h, 1) - f(1, 1)) / -h in_y = (f(1, 1 + h) - f(1, 1)) / h, (f(1, 1 - h) - f(1, 1)) / -h print(h, [round(v, 6) for v in in_x], [round(v, 6) for v in in_y])

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The rules of Level 2 still work

To find $f_x$ , treat $y$ as a fixed number, such as 5, and differentiate in $x$ by the usual rules. To find $f_y$ , do the reverse. Nothing new is needed.

A polynomial. Take $f(x, y) = x^2y + 3y^2$ . In $x^2y$ the $y$ is a constant factor, so its $x$ -derivative is $2x \cdot y$ (lesson 2.3.1). The term $3y^2$ does not contain $x$ at all, so it is a constant and its $x$ -derivative is 0.

\begin{gathered} f_x = 2xy \\[4pt] f_y = x^2 + 6y \end{gathered}

In $f_y$ , the $x^2$ is a constant factor of $y$ , which has slope 1. At $(2, 1)$ the function is 7, with $f_x = 4$ and $f_y = 10$ .

The chain rule. For $g(x, y) = e^{xy}$ , the inside is $xy$ . Its slope in $x$ is $y$ , so $g_x = ye^{xy}$ (lessons 2.3.3 and 2.3.4).

In the same way $g_y = xe^{xy}$ . At $(1, 2)$ : $g_x = 2e^2 \approx 14.7781$ and $g_y = e^2 \approx 7.3891$ .

The quotient rule, or a power. For $q(x, y) = \frac{x}{y}$ , the $\frac{1}{y}$ is a constant factor in $x$ , so $q_x = \frac{1}{y}$ . In $y$ , write $q = x\,y^{-1}$ : then $q_y = -\frac{x}{y^2}$ . At $(3, 2)$ : $q_x = \frac{1}{2}$ and $q_y = -\frac{3}{4}$ .

Rule: Computing a partial derivative

\begin{gathered} f_x: \text{ treat } y \text{ as a constant} \\[4pt] f_y: \text{ treat } x \text{ as a constant} \end{gathered}

Then use the rules of lessons 2.3.1 to 2.3.5. A term without the moving variable has derivative 0; a factor without it stays as a factor.

x

import sympy as sp x, y = sp.symbols("x y") for f, point in [(x**2 * y + 3 * y**2, {x: 2, y: 1}), (sp.exp(x * y), {x: 1, y: 2}), (x / y, {x: 3, y: 2})]: fx, fy = sp.diff(f, x), sp.diff(f, y) print(f, "|", fx, "|", fy, "|", fx.subs(point), fy.subs(point)) print(" at the point:", [round(float(v.subs(point)), 4) for v in (f, fx, fy)])

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Signs, units and where it breaks

The sign of $f_x$ says what happens when $x$ grows and $y$ stays put. Positive means $f$ rises, negative means it falls, and zero means the slice is flat there. The other partial can tell a different story at the same point.

For the saddle $x^2 - y^2$ of lesson 4.1.1, $f_x = 2x$ and $f_y = -2y$ . At $(1, 1)$ , $f_x = 2$ and $f_y = -2$ : uphill along $x$ , downhill along $y$ . At $(0, 0)$ both are 0, and for the plane $6 - 2x - 3y$ they are $-2$ and $-3$ everywhere.

Units

A partial derivative is in output units per unit of the input that moves. A tin can of radius $r$ cm and height $\ell$ cm holds $V = \pi r^2\ell$ cm³. Then $\frac{\partial V}{\partial r} = 2\pi r\ell$ is in cm³ per cm of radius, and $\frac{\partial V}{\partial \ell} = \pi r^2$ in cm³ per cm of height.

For $r = 3$ and $\ell = 10$ , $V = 282.74$ cm³, $\frac{\partial V}{\partial r} = 188.5$ and $\frac{\partial V}{\partial \ell} = 28.27$ . Widening the can is far more effective than making it taller: per centimetre, more than six times as many cm³.

Each partial gives a tangent estimate in its own direction, as in lesson 2.2.1. With $\ell$ fixed at 10, widening $r$ by 0.1 adds about $188.5 \times 0.1 = 18.85$ cm³; the true gain is 19.16. With $r$ fixed at 3, adding 0.1 to $\ell$ adds 2.83 cm³, exactly, because $V$ is a straight line in $\ell$ .

Where a partial derivative does not exist

A partial derivative is a one-variable derivative, so it fails where lesson 2.2.2's did. Take $g(x, y) = |x| + y^2$ at $(0, 1)$ .

With $y$ fixed, the slice $|x| + 1$ has a corner at 0. Its chord slopes are 1 from the right and $-1$ from the left, so $g_x$ does not exist there.

The other slice is smooth. With $x = 0$ fixed, the slice is $g(0, y) = y^2$ , so $g_y = 2y = 2$ at $(0, 1)$ . One partial can exist where the other does not.

|x| + y^2

import math def volume(r, height): return math.pi * r**2 * height r, height = 3, 10 print(round(volume(r, height), 2), round(2 * math.pi * r * height, 2), round(math.pi * r**2, 2)) print(round(0.1 * 2 * math.pi * r * height, 2), round(volume(3.1, height) - volume(r, height), 2), round(volume(r, 10.1) - volume(r, height), 2)) def g(x, y): return abs(x) + y**2 for step in [0.1, 0.001]: print(step, round((g(step, 1) - g(0, 1)) / step, 6), round((g(-step, 1) - g(0, 1)) / -step, 6), round((g(0, 1 + step) - g(0, 1)) / step, 6))

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Second partial derivatives

$f_x$ and $f_y$ are functions of $x$ and $y$ too, so each has two partial derivatives of its own. That gives four second partial derivatives:

\begin{gathered} f_{xx} = (f_x)_x, \quad f_{yy} = (f_y)_y \\[4pt] f_{xy} = (f_x)_y, \quad f_{yx} = (f_y)_x \end{gathered}

$f_{xx}$ is the second derivative of the slice in $x$ : it measures how that slice bends (lesson 2.4.2). For the bowl, $f_{xx} = 2$ and $f_{yy} = 4$ , so the slices in $y$ bend twice as sharply.

The mixed partials $f_{xy}$ and $f_{yx}$ are new. $f_{xy}$ says how the slope in $x$ changes as $y$ moves.

A worked example

Take $f(x, y) = x^2y + 3y^2$ again, with $f_x = 2xy$ . At $x = 2$ that slope is $4y$ . Read it off three slices:

slice	$y = 0$	$y = 1$	$y = 2$
$f(x, y)$ as a function of $x$	0	$x^2 + 3$	$2x^2 + 12$
its slope at $x = 2$ , $f_x(2, y)$	0	4	8

Each step of 1 in $y$ adds 4 to the slope. So $f_{xy}(2, y) = 4$ . Differentiating $f_x = 2xy$ in $y$ gives the same: $f_{xy} = 2x$ , which is 4 at $x = 2$ .

The slope in $x$ at $x = 2$ grows as $y$ grows: 0, 4 and 8 on the slices $y = 0$ , 1 and 2. Its rate of growth, 4 per unit of $y$ , is the mixed partial derivative.

Now go the other way: $f_y = x^2 + 6y$ , and its $x$ -derivative is $f_{yx} = 2x$ . The two mixed partials agree. The rest: $f_{xx} = 2y$ and $f_{yy} = 6$ , so at $(2, 1)$ they are 2 and 6.

Rule: Mixed partials agree

f_{xy} = f_{yx}

The order of differentiation does not matter when the second partials are continuous, which holds for every function in this course. This lesson states it; the Proof Gym proves it.

So a function of two variables has only three different second partials: $f_{xx}$ , $f_{xy}$ and $f_{yy}$ . Lesson 4.3.1 puts them in a symmetric matrix, the Hessian $H$ .

Two notations need care. $f_{xy}$ is read left to right: first $x$ , then $y$ .

The same thing written with $\partial$ is $\frac{\partial^2 f}{\partial y\,\partial x}$ , read right to left. Since the two agree, the order rarely changes an answer.

f_{xx}

import sympy as sp x, y = sp.symbols("x y") for f in [x**2 * y + 3 * y**2, sp.exp(x * y), x**2 + 2 * y**2, x * y]: fxx = sp.diff(f, x, 2) fxy = sp.diff(f, x, y) # x first, then y fyx = sp.diff(f, y, x) # y first, then x fyy = sp.diff(f, y, 2) print(f, "|", fxx, "|", sp.factor(fxy), "|", sp.simplify(fxy - fyx) == 0, "|", fyy) fx = sp.diff(x**2 * y + 3 * y**2, x) print([fx.subs({x: 2, y: b}) for b in [0, 1, 2]])

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For the bowl the mixed partial is 0: its slope in $x$ , $2x$ , does not depend on $y$ . For $xy$ it is 1, since the slice $y = b$ has slope $b$ .

Estimating partials by central differences

When $f$ is a program rather than a formula, estimate each partial by lesson 2.8.1's central difference. Move one input up and down by $h$ , and keep the other exactly where it was.

Rule: Central differences in two variables

\begin{gathered} f_x \approx \frac{f(x + h, y) - f(x - h, y)}{2h} \\[6pt] f_{xx} \approx \\ \frac{f(x + h, y) - 2f(x, y) + f(x - h, y)}{h^2} \\[6pt] f_{xy} \approx \frac{f_{++} - f_{+-} - f_{-+} + f_{--}}{4h^2} \end{gathered}

$f_y$ and $f_{yy}$ are the same with $y$ moving. In the last line $f_{+-}$ means $f(x + h, y - h)$ , and so on. It is the central difference in $y$ of the central difference in $x$ , from four corner points.

For $g = e^{xy}$ at $(1, 2)$ the true values are $g_x = 2e^2 \approx 14.778112$ and $g_{xy} = (1 + xy)e^{xy} = 3e^2 \approx 22.167168$ .

step $h$	0.1	0.01	0.001
estimate of $g_x$	14.87683	14.779097	14.778122
its error	$9.9 \times 10^{-2}$	$9.9 \times 10^{-4}$	$9.9 \times 10^{-6}$
estimate of $g_{xy}$	22.477557	22.170247	22.167199
its error	$3.1 \times 10^{-1}$	$3.1 \times 10^{-3}$	$3.1 \times 10^{-5}$

Both errors shrink like $h^2$ : ten times smaller $h$ , a hundred times smaller error. The first matches lesson 2.8.1's $\frac{1}{6}g_{xxx}h^2$ , with $g_{xxx} = y^3e^{xy} = 8e^2$ here.

Rounding sets a floor, as in lesson 2.8.1. For a first partial a step near $10^{-5}$ times the scale of the input works well, and for second partials near $10^{-4}$ . With $h = 10^{-8}$ the mixed estimate is ruined by rounding.

h = 10^{-4}

import math def g(x, y): return math.exp(x * y) e2 = math.exp(2) exact = 2 * e2, 3 * e2 # g_x and g_xy at (1, 2) def central_x(f, x, y, h): return (f(x + h, y) - f(x - h, y)) / (2 * h) def mixed(f, x, y, h): return (f(x + h, y + h) - f(x + h, y - h) - f(x - h, y + h) + f(x - h, y - h)) / (4 * h**2) print(round(exact[0], 6), round(exact[1], 6)) for h in [0.1, 0.01, 0.001]: gx, gxy = central_x(g, 1, 2, h), mixed(g, 1, 2, h) print(h, round(gx, 6), f"{gx - exact[0]:.1e}", round(gxy, 6), f"{gxy - exact[1]:.1e}") print(abs(mixed(g, 1, 2, 1e-4) - exact[1]) < 1e-5, abs(mixed(g, 1, 2, 1e-8) - exact[1]) > 1e-3)

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Three or more variables

Nothing changes with more inputs: to differentiate in one, hold all the others fixed. A box with sides $x$ , $y$ and $z$ cm has volume $V = xyz$ cm³, with

V_x = yz, \quad V_y = xz, \quad V_z = xy

For a box of $2 \times 3 \times 4$ cm, $V = 24$ and the partials are 12, 8 and 6 cm³ per cm. The shortest side gives the most: adding 0.1 cm to $x$ adds exactly $12 \times 0.1 = 1.2$ cm³, since $V$ is a straight line in $x$ .

There are now nine second partials: three on the diagonal, such as $V_{xx}$ , and three equal pairs, such as $V_{xy} = V_{yx} = z$ . For the box they are 4, 3 and 2 off the diagonal, and $V_{xx} = V_{yy} = V_{zz} = 0$ .

xyz

import sympy as sp x, y, z = sp.symbols("x y z") V = x * y * z point = {x: 2, y: 3, z: 4} print([sp.diff(V, v).subs(point) for v in (x, y, z)]) print(sp.Matrix(3, 3, lambda i, j: sp.diff(V, (x, y, z)[i], (x, y, z)[j]).subs(point))) print(round(float(V.subs({x: 2.1, y: 3, z: 4}) - V.subs(point)), 6))

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Case study 4.1.4 treats an option's value as a function of the stock price, the volatility and time. Its sensitivities are partial derivatives. Lesson 4.2.1 puts the first partials together into one vector, the gradient.

Common mistakes

The fixed variable is a number, not a variable. $\frac{\partial}{\partial x}(x^2y) = 2xy$ . Using the product rule as if $y$ moved gives an extra term that does not belong.
A term without the moving variable vanishes; a factor stays. $\frac{\partial}{\partial x}(y^2) = 0$ , but $\frac{\partial}{\partial x}(xy^2) = y^2$ .
A value needs both coordinates. For the bowl $f_x = 2x$ does not involve $y$ , but the point is still $(1, 1)$ , where $f_x = 2$ and $f_y = 4$ .
Units are per unit of the moving input. $\frac{\partial V}{\partial r}$ for the can is in cm³ per cm of radius.
One partial can exist where another does not. For $|x| + y^2$ at $(0, 1)$ , $g_y = 2$ but $g_x$ does not exist.
Mind the order in the notation. $f_{xy}$ means $x$ first; $\frac{\partial^2 f}{\partial y\,\partial x}$ means the same.
In a central difference, move only one input, and keep the other exactly fixed. A mixed partial needs four corner points.

Partial derivatives

0 of 13 correct

1.
For $f(x, y) = x^2 + 2y^2$ , find $f_y(1, -2)$
2.
For $f(x, y) = x^3y^2$ , find $f_x$
3.
For $f(x, y) = x^3y^2$ , find $f_y$
4.
For $f(x, y) = 5x + y^4$ , find $f_x$
5.
For $g(x, y) = e^{2x - y}$ , find $g_y$ at $(1, 2)$
6.
For $q(x, y) = \frac{x}{y}$ , find $q_y$ at $(4, 2)$
7.
For the saddle $x^2 - y^2$ , find $f_y$ at $(3, 2)$
8.
A tin can holds $V = \pi r^2 \ell$ cm³. Find $\frac{\partial V}{\partial \ell}$ at $r = 2$ , in cm³ per cm (type pi)
9.
For $f(x, y) = x^2y + 3y^2$ , find $f_{xy}$ at $(5, 1)$
10.
For $f(x, y) = x^3y^2$ , find $f_{yx}$
11.
For $f(x, y) = x^2 + 2y^2$ , find $f_{yy}$
12.
The central difference for $f_x$ of $f(x, y) = x^2y$ at $(1, 3)$ with $h = 0.1$ is $\frac{f(1.1, 3) - f(0.9, 3)}{0.2}$ . Find it
13.
A box has $V = xyz$ . Find $V_z$ at $(2, 3, 4)$

How to type answers

Powers with ^: x^7, 2^(1/3), a^-2.
Multiply with * or by writing side by side: 3x^2, 2(x+1).
ln(x) is the natural log, log(x, 2) is log base 2, e is Euler's number, sqrt(x) is a square root.
sin(x), cos(x) and tan(x) take radians; write pi for π and sin(x)^2 for sin²x.
Your first answer to each question is saved: it counts towards your score and brings the question back for review.
Exact answers like ln(20)/ln(3) are best. Decimals need about 4 significant figures.

\frac{\partial f}{\partial x}

Partial derivatives

The slope of a slice

Rule: Partial derivatives

Notation

The rules of Level 2 still work

Rule: Computing a partial derivative

Signs, units and where it breaks

Units

Where a partial derivative does not exist

Second partial derivatives

A worked example

Rule: Mixed partials agree

Estimating partials by central differences

Rule: Central differences in two variables

Three or more variables

Common mistakes

Partial derivatives

Summary card

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