S1 CSE Maths · Module 2

Page 1 of 13

Module 2 at a glance

Module 1 was about functions of one variable. But area, volume, temperature on a metal plate and the height of a hill depend on two or more inputs. Module 2 takes the same story to more variables, in the same order: limits → derivatives → using derivatives.

The only new difficulty is that with two inputs you can move in many directions. So every idea has one extra question: in which direction?

Functions of two variables $z=f(x,y)$ : what they are and where they are defined.
Limits and continuity: now we can reach a point along infinitely many paths, not just from the left and the right.
Partial derivatives: change one input and keep the others fixed.
Chain rule and implicit differentiation: how a change passes through a chain of dependent variables.
Applications: slopes of surfaces, related rates, and level curves.

What you will be able to do

State the domain and range of a function of two variables.
Find a two-variable limit, and prove that a limit does not exist by comparing two paths.
Check whether a function is continuous at a point.
Find first, second and mixed partial derivatives, and check that $f_{xy}=f_{yx}$ .
Use the chain rule with one or two parameters, and draw the tree diagram that helps you organise it.
Find $\dfrac{dy}{dx}$ for an implicit curve using partial derivatives.
Solve related-rates problems with many changing quantities.
Sketch the level curves of a function.

How to use these notes

Every topic has the same shape: idea → rule → worked example → common mistakes.
Try each worked example yourself first. Cover the answer, solve it, and then compare.
Green boxes hold key ideas. Amber boxes marked Watch out show the mistakes that make you lose marks.
The last page has practice questions with answers, and then a one-page formula sheet for revision.

Where each topic sits

Part	Topics
Functions	Domain, range
Limits and continuity	Limits along paths, the two-path test, continuity
Partial derivatives	First order, slope of a surface, second order, mixed partials
Chain rule	One parameter, two parameters, related rates
Implicit functions	$dy/dx=-f_x/f_y$
Pictures	Level curves

Before you start. You should know the differentiation from Module 1: the power, product, quotient and chain rules, and the derivatives of $e^x$ , $\ln x$ , $\sin x$ , $\cos x$ , $\tan x$ and $\tan^{-1}x$ . Partial derivatives are just ordinary derivatives with a little extra care, so these rules do all the work.

Page 2 of 13

Functions of two variables

The idea

A function of two variables gives exactly one number for each pair $(x,y)$ in its domain:

z=f(x,y)

Examples you already know: the area of a rectangle $A=xy$ , the volume of a cone $V=\tfrac13\pi r^2h$ , the distance from the origin $\sqrt{x^2+y^2}$ .

The graph of $z=f(x,y)$ is a surface in space: above each point $(x,y)$ of the plane, there is a point at height $z$ .

Domain and range

The domain is the set of all $(x,y)$ for which $f(x,y)$ is defined. It is a region of the plane, not an interval.
The range is the set of all values $f(x,y)$ can take.

The conditions are the same as in one variable:

Trouble	Rule
Division	the denominator must not be $0$
Even root	the expression under it must be $\ge 0$
Logarithm	the expression inside must be $>0$

Example 1. $f(x,y)=x^2-y^2$ .

There is no problem anywhere, so the domain is all of $\mathbb{R}^2$ . Every real number is a value: to get $k\ge0$ take $(\sqrt k,\,0)$ , and to get $k<0$ take $(0,\,\sqrt{-k})$ . So the range is $\mathbb{R}$ .

Example 2. $f(x,y)=\dfrac{x+y}{x-y}$ .

The denominator becomes zero when $x=y$ . The domain is the whole plane except the line $y=x$ :

\mathbb{R}^2\setminus\{(x,y): x=y\}

Every real $k$ is a value: for $k\ne-1$ take $x=1,\ y=\dfrac{k-1}{k+1}$ , and for $k=-1$ take $(0,1)$ . So the range is $\mathbb{R}$ .

Example 3. $f(x,y)=\sqrt{9-x^2-y^2}$ .

The root needs $9-x^2-y^2\ge0$ , that is $x^2+y^2\le9$ . The domain is the closed disc of radius $3$ centred at the origin. The largest value is $f(0,0)=3$ and the smallest is $0$ (on the edge of the disc), so the range is $[0,3]$ . This is the upper half of a sphere. You will see it again in level curves.

Watch out. Write the domain as an equation or inequality in $x$ and $y$ (like " $x\neq y$ " or " $x^2+y^2\le9$ "). Writing only a one-variable condition like " $x\neq0$ " is not enough, because the condition usually involves both variables together.

Page 3 of 13

Limits of functions of two variables

The idea

\lim_{(x,y)\to(a,b)}f(x,y)=L

means: $f(x,y)$ comes as close to $L$ as we like when the point $(x,y)$ is close enough to $(a,b)$ , but not equal to it.

On a number line you can reach a point from only two sides, left and right. In the plane you can reach $(a,b)$ along infinitely many paths: straight lines, parabolas, spirals.

Key idea. The limit exists only if $f$ goes to the same number along every path to $(a,b)$ . If even one path gives a different answer, the limit does not exist.

Easy case: just substitute

The limit laws from Module 1 work in the same way for $(x,y)\to(a,b)$ : you can take limits through sums, differences, constant multiples, products, quotients (when the limit of the denominator is $\ne0$ ), powers and roots.

So for a polynomial, or a rational function whose denominator is not $0$ at the point, substitute.

Example 1. $\displaystyle\lim_{(x,y)\to(0,1)}\frac{x-xy+3}{x^2y+5xy-y^3}$

\frac{0-0\cdot1+3}{0^2\cdot1+5\cdot0\cdot1-1^3}=\frac{3}{-1}=\mathbf{-3}

Example 2. $\displaystyle\lim_{(x,y)\to(3,-4)}\sqrt{x^2+y^2}$

\sqrt{3^2+(-4)^2}=\sqrt{9+16}=\sqrt{25}=\mathbf{5}

This is the distance from $(3,-4)$ to the origin.

Example 3. $\displaystyle\lim_{(x,y)\to(\pi/2,\,0)}\left(\frac{x}{\sin x}-\frac{\sin y}{y}\right)$

Take the two terms separately. The first is $\dfrac{\pi/2}{\sin(\pi/2)}=\dfrac{\pi}{2}$ . The second gives $0/0$ on substitution, but it is the standard limit $\dfrac{\sin y}{y}\to1$ from Module 1. So the answer is

\frac{\pi}{2}-1

When substitution gives $0/0$

Do the same as in Module 1: factorise and cancel, or use the difference of squares.

Example 4. $\displaystyle\lim_{(x,y)\to(0,0)}\frac{x^2-xy}{\sqrt{x}-\sqrt{y}}$

Substituting gives $0/0$ . Factorise the top as $x(x-y)$ and write $x-y=(\sqrt x-\sqrt y)(\sqrt x+\sqrt y)$ :

\frac{x(x-y)}{\sqrt x-\sqrt y}=\frac{x(\sqrt x-\sqrt y)(\sqrt x+\sqrt y)}{\sqrt x-\sqrt y}=x(\sqrt x+\sqrt y)

Now substitute: $0\cdot(0+0)=\mathbf{0}$ . (The square roots need $x,y\ge0$ , and $x\ne y$ keeps the original fraction defined.)

Showing a limit does not exist: the two-path test

Two-path test. If $f$ goes to different values along two different paths to $(a,b)$ , then $\displaystyle\lim_{(x,y)\to(a,b)}f(x,y)$ does not exist.

For the origin, the usual choice is the family of lines through it, $y=mx$ . Put $y=mx$ , let $x\to0$ , and see whether the answer depends on $m$ . If it does, the limit does not exist.

Example 5. Show that $\displaystyle\lim_{(x,y)\to(0,0)}\frac{x^2-y^2}{x^2+y^2}$ does not exist.

Substitution gives $0/0$ , so the limit laws do not help. Put $y=mx$ :

\lim_{x\to0}\frac{x^2-m^2x^2}{x^2+m^2x^2}=\lim_{x\to0}\frac{x^2(1-m^2)}{x^2(1+m^2)}=\frac{1-m^2}{1+m^2}

The answer depends on $m$ . Along the $x$ -axis ( $m=0$ ) it is $1$ , and along $y=x$ ( $m=1$ ) it is $0$ . Different paths give different values, so the limit does not exist.

Example 6. Show that $\displaystyle\lim_{(x,y)\to(0,0)}\frac{xy}{x^2+y^2}$ does not exist.

Put $y=mx$ :

\lim_{x\to0}\frac{x\cdot mx}{x^2+m^2x^2}=\frac{m}{1+m^2}

This is $0$ for $m=0$ and $\tfrac12$ for $m=1$ . The limit does not exist.

Watch out. The two-path test can only prove that a limit does not exist. If every straight line gives the same value, that does not prove that the limit exists, because some curve may give a different value. For example $\dfrac{xy^2}{x^2+y^4}$ gives $0$ along every line $y=mx$ , but along the parabola $x=y^2$ it equals $\dfrac{y^4}{2y^4}=\dfrac12$ . So its limit at the origin does not exist.

Page 4 of 13

Continuity

The idea

A function is continuous at a point when its graph has no break there: the value we expect from the nearby points is the value the function actually has.

Definition. $f(x,y)$ is continuous at $(x_0,y_0)$ when all three hold:

$f(x_0,y_0)$ is defined.

$\displaystyle\lim_{(x,y)\to(x_0,y_0)}f(x,y)$ exists.

$\displaystyle\lim_{(x,y)\to(x_0,y_0)}f(x,y)=f(x_0,y_0)$ .

$f$ is continuous if it is continuous at every point of its domain.

What you can take for granted

Polynomials are continuous everywhere.
A rational function is continuous everywhere its denominator is not $0$ .
Sums, products, quotients (denominator $\ne0$ ) and compositions of continuous functions are continuous.

So a problem can come only where the formula changes (a piecewise function), or where a denominator becomes zero. These are the points to check.

Method

Away from the doubtful point, the function is made of continuous parts, so it is continuous there.
At the doubtful point, check the three conditions. Most of the time, condition 2 fails. Use the two-path test.

Examples

Example 1. Show that

f(x,y)=\begin{cases}\dfrac{2xy}{x^2+y^2}, & (x,y)\neq(0,0)\\[8pt] 0, & (x,y)=(0,0)\end{cases}

is continuous everywhere except the origin.

Away from the origin it is a rational function with a non-zero denominator, so it is continuous.

At the origin $f(0,0)=0$ is defined. Along $y=mx$ :

\lim_{x\to0}\frac{2x\cdot mx}{x^2+m^2x^2}=\frac{2m}{1+m^2}

This depends on $m$ , so the limit does not exist and $f$ is not continuous at the origin. Changing the value of $f(0,0)$ cannot fix this, because there is no single limit to match.

Example 2. Show that

f(x,y)=\begin{cases}\dfrac{y}{\sqrt{x^2+y^2}}, & (x,y)\neq(0,0)\\[8pt] 0, & (x,y)=(0,0)\end{cases}

is discontinuous at $(0,0)$ .

Along $y=mx$ with $x>0$ we have $\sqrt{x^2+m^2x^2}=x\sqrt{1+m^2}$ , so

\lim_{x\to0^+}\frac{mx}{x\sqrt{1+m^2}}=\frac{m}{\sqrt{1+m^2}}

This depends on $m$ (it is $0$ for $m=0$ and $\tfrac1{\sqrt2}$ for $m=1$ ), so the limit does not exist and $f$ is discontinuous at $(0,0)$ .

Example 3. Let $f(x,y)=x^2+y^2$ for $(x,y)\neq(0,0)$ , and $f(0,0)=1$ . Is $f$ continuous at the origin?

Conditions 1 and 2 hold: $f(0,0)=1$ is defined, and the limit is $0^2+0^2=0$ . But condition 3 fails because $0\neq1$ . So $f$ is not continuous at the origin. This is a removable break: if we redefine $f(0,0)=0$ , it is fixed.

Watch out. "The limit exists" and "the function is continuous" are different things. Example 3 has a limit but is still discontinuous. All three conditions must be true.

Page 5 of 13

Partial derivatives

The idea

For $z=f(x,y)$ there are two natural questions: how fast does $z$ change if we move in the $x$ direction (keeping $y$ fixed), and how fast does it change if we move in the $y$ direction (keeping $x$ fixed)? The answers are the partial derivatives.

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

Each one is just the Module 1 definition of a derivative, applied to one variable while the other is kept fixed.

Name	Common symbols
partial derivative with respect to $x$	$\dfrac{\partial z}{\partial x},\quad \dfrac{\partial f}{\partial x},\quad f_x$
partial derivative with respect to $y$	$\dfrac{\partial z}{\partial y},\quad \dfrac{\partial f}{\partial y},\quad f_y$

The curly $\partial$ means "there are other variables also". The straight $d$ is only for functions of one variable.

Method

Steps.

For $f_x$ : treat $y$ as a constant and differentiate with respect to $x$ using the usual rules.

For $f_y$ : treat $x$ as a constant and differentiate with respect to $y$ .

A constant multiplier stays as it is. A term with only the other variable is a constant, so it becomes zero. For example, for $x^2y$ : differentiating in $x$ gives $2xy$ (the $y$ just stays as a multiplier), and differentiating in $y$ gives $x^2$ .

Examples

Example 1. $f(x,y)=x^2y+3xy^2$

f_x=2xy+3y^2\qquad f_y=x^2+6xy

Example 2. $f(x,y)=2x^2-3y-4$

f_x=4x\qquad f_y=-3

The $-3y$ is constant with respect to $x$ , so it disappears from $f_x$ . The $2x^2$ and $-4$ disappear from $f_y$ .

Example 3. $f(x,y)=(x+1)(y-2)$ . Treat the other bracket as a constant factor.

f_x=y-2\qquad f_y=x+1

Example 4. $f(x,y)=\sqrt{x^2+y^2}$ . Use the chain rule on the root.

f_x=\frac{1}{2\sqrt{x^2+y^2}}\cdot2x=\frac{x}{\sqrt{x^2+y^2}}\qquad f_y=\frac{y}{\sqrt{x^2+y^2}}

Example 5. $z=\sqrt{3x+4y}$

\frac{\partial z}{\partial x}=\frac{1}{2\sqrt{3x+4y}}\cdot3=\frac{3}{2\sqrt{3x+4y}}\qquad \frac{\partial z}{\partial y}=\frac{1}{2\sqrt{3x+4y}}\cdot4=\frac{2}{\sqrt{3x+4y}}

Example 6. $f(x,y)=\log(x^2+y^2)$ (natural logarithm)

f_x=\frac{1}{x^2+y^2}\cdot2x=\frac{2x}{x^2+y^2}\qquad f_y=\frac{2y}{x^2+y^2}

Example 7. $f(x,y)=e^{ax}\sin by$ . Here $a$ and $b$ are constants.

f_x=a\,e^{ax}\sin by\qquad f_y=b\,e^{ax}\cos by

Example 8. $f(x,y)=x\,e^{-y}+5y$

f_x=e^{-y}\qquad f_y=x\cdot e^{-y}\cdot(-1)+5=5-x\,e^{-y}

Example 9. $f(x,y)=e^{x^2y^3}\sin(x^2-y^2)$ . Use the product rule, and the chain rule inside each factor.

\begin{aligned} f_x&=e^{x^2y^3}\cos(x^2-y^2)\cdot2x+\sin(x^2-y^2)\cdot e^{x^2y^3}\cdot2xy^3\\ &=2x\,e^{x^2y^3}\cos(x^2-y^2)+2xy^3e^{x^2y^3}\sin(x^2-y^2) \end{aligned}

\begin{aligned} f_y&=e^{x^2y^3}\cos(x^2-y^2)\cdot(-2y)+\sin(x^2-y^2)\cdot e^{x^2y^3}\cdot3x^2y^2\\ &=-2y\,e^{x^2y^3}\cos(x^2-y^2)+3x^2y^2e^{x^2y^3}\sin(x^2-y^2) \end{aligned}

Example 10. $f(x,y)=\dfrac{x-y}{x^2+y^2}$ . Use the quotient rule.

f_x=\frac{(x^2+y^2)\cdot1-(x-y)\cdot2x}{(x^2+y^2)^2}=\frac{y^2-x^2+2xy}{(x^2+y^2)^2}

f_y=\frac{(x^2+y^2)\cdot(-1)-(x-y)\cdot2y}{(x^2+y^2)^2}=\frac{y^2-x^2-2xy}{(x^2+y^2)^2}

Three or more variables

Nothing changes. For $u=f(x,y,z)$ , to find $u_x$ treat both $y$ and $z$ as constants, and similarly for the others.

Watch out. Students often forget the chain-rule factors. In Example 5 the $3$ and the $4$ come from differentiating the inside part $3x+4y$ . In Example 8 the $-1$ comes from differentiating $-y$ . Also, never write $d$ where you should write $\partial$ .

Page 6 of 13

Slope and rate of change

The idea

Take the surface $z=f(x,y)$ and a point $(a,b)$ in the plane.

Cut the surface with the vertical plane $y=b$ . You get an ordinary curve $z=f(x,b)$ . Its slope at $x=a$ is $f_x(a,b)$ .
Cut with the plane $x=a$ . The slope of that curve at $y=b$ is $f_y(a,b)$ .

Meaning. $\dfrac{\partial z}{\partial x}$ is the slope of the surface in the $x$ -direction, and also the rate of change of $z$ with respect to $x$ while $y$ is held fixed. Likewise $\dfrac{\partial z}{\partial y}$ for the $y$ -direction.

To find the value at a point, differentiate first and then substitute. The notation for "value at $(a,b)$ " is

\left.\frac{\partial z}{\partial x}\right|_{(a,b)}

The sign tells the story. A positive value means $z$ is increasing as you move in the positive direction. A negative value means it is decreasing. Zero means the surface is flat in that direction.

Examples

Example 1. Find the slope of the surface $z=10-4x^2-2y^2$ at $(1,2)$ in the $x$ -direction and the $y$ -direction.

\frac{\partial z}{\partial x}=-8x\ \Rightarrow\ \left.\frac{\partial z}{\partial x}\right|_{(1,2)}=-8(1)=\mathbf{-8}

\frac{\partial z}{\partial y}=-4y\ \Rightarrow\ \left.\frac{\partial z}{\partial y}\right|_{(1,2)}=-4(2)=\mathbf{-8}

The surface is a dome with its top at $(0,0,10)$ . At $(1,2)$ it goes down with the same steepness in both directions.

Example 2. Find the rate of change of $z=\sin(y^2-4x)$ with respect to $y$ at $(3,1)$ .

\frac{\partial z}{\partial y}=\cos(y^2-4x)\cdot2y=2y\cos(y^2-4x)

\left.\frac{\partial z}{\partial y}\right|_{(3,1)}=2(1)\cos(1-12)=2\cos(-11)=\mathbf{2\cos 11}

Cosine is an even function, so $\cos(-11)=\cos 11$ . In radians this is about $0.009$ , so at that point $z$ is almost flat in the $y$ -direction.

Watch out. Put the point after differentiating. If you put $y=1$ in $z$ first, then $y$ becomes a constant and differentiating with respect to $y$ just gives $0$ .

Page 7 of 13

Second-order partial derivatives

The idea

A partial derivative is also a function of $x$ and $y$ , so we can differentiate it again. A function of two variables has four second-order partial derivatives.

Symbol	Means	In words
$f_{xx}=\dfrac{\partial^2f}{\partial x^2}$	$\dfrac{\partial}{\partial x}\!\left(\dfrac{\partial f}{\partial x}\right)$	differentiate with respect to $x$ twice
$f_{yy}=\dfrac{\partial^2f}{\partial y^2}$	$\dfrac{\partial}{\partial y}\!\left(\dfrac{\partial f}{\partial y}\right)$	differentiate with respect to $y$ twice
$f_{xy}=\dfrac{\partial^2f}{\partial y\,\partial x}$	$\dfrac{\partial}{\partial y}\!\left(\dfrac{\partial f}{\partial x}\right)$	$x$ first, then $y$
$f_{yx}=\dfrac{\partial^2f}{\partial x\,\partial y}$	$\dfrac{\partial}{\partial x}\!\left(\dfrac{\partial f}{\partial y}\right)$	$y$ first, then $x$

The last two are the mixed derivatives. The subscripts in $f_{xy}$ are read left to right, but the $\partial$ notation is read right to left. Some books swap the names. This is not a problem here because of the next fact.

Mixed partials are equal. If $f_{xy}$ and $f_{yx}$ are continuous, then $f_{xy}=f_{yx}$ . Every function in this module satisfies this, so the order of differentiation does not matter. Use it to check your answer: find both mixed derivatives and see that they are equal.

Examples

Example 1. Find all second-order partial derivatives of $f(x,y)=x^3y+2xy^2$ .

First order:

f_x=3x^2y+2y^2\qquad f_y=x^3+4xy

Second order:

f_{xx}=6xy\qquad f_{yy}=4x

\frac{\partial^2f}{\partial y\,\partial x}=\frac{\partial}{\partial y}(3x^2y+2y^2)=3x^2+4y \qquad \frac{\partial^2f}{\partial x\,\partial y}=\frac{\partial}{\partial x}(x^3+4xy)=3x^2+4y

The mixed derivatives agree.

Example 2. For $f(x,y)=2x^5y^3+5x+7y$ find the mixed derivatives.

f_x=10x^4y^3+5\qquad f_y=6x^5y^2+7

\frac{\partial^2f}{\partial y\,\partial x}=30x^4y^2\qquad\frac{\partial^2f}{\partial x\,\partial y}=30x^4y^2

(Also $f_{xx}=40x^3y^3$ and $f_{yy}=12x^5y$ .)

Example 3. For $f(x,y)=x^3y+e^{xy^2}$ show that $\dfrac{\partial^2f}{\partial y\,\partial x}=\dfrac{\partial^2f}{\partial x\,\partial y}$ .

f_x=3x^2y+y^2e^{xy^2}\qquad f_y=x^3+2xy\,e^{xy^2}

Differentiate $f_x$ with respect to $y$ (product rule on the second term):

\frac{\partial^2f}{\partial y\,\partial x}=3x^2+2y\,e^{xy^2}+y^2\cdot e^{xy^2}\cdot2xy=3x^2+2y\,e^{xy^2}+2xy^3e^{xy^2}

Differentiate $f_y$ with respect to $x$ :

\frac{\partial^2f}{\partial x\,\partial y}=3x^2+2y\,e^{xy^2}+2xy\cdot e^{xy^2}\cdot y^2=3x^2+2y\,e^{xy^2}+2xy^3e^{xy^2}

They are equal. $\checkmark$

Third order

Derivatives of any order work in the same way. In $\dfrac{\partial^3f}{\partial x\,\partial y\,\partial z}$ read from right to left: first $z$ , then $y$ , then $x$ .

Example 4. If $f(x,y,z)=e^{xyz}$ , find $\dfrac{\partial^3f}{\partial x\,\partial y\,\partial z}$ .

\frac{\partial f}{\partial z}=xy\,e^{xyz}

\frac{\partial^2f}{\partial y\,\partial z}=x\,e^{xyz}+xy\cdot xz\,e^{xyz}=(x+x^2yz)\,e^{xyz}

\begin{aligned} \frac{\partial^3f}{\partial x\,\partial y\,\partial z}&=(1+2xyz)\,e^{xyz}+(x+x^2yz)\,yz\,e^{xyz}\\ &=e^{xyz}\left(1+2xyz+xyz+x^2y^2z^2\right)=\mathbf{e^{xyz}\left(1+3xyz+x^2y^2z^2\right)} \end{aligned}

Proving an equation with derivatives

Example 5. Laplace's equation. If $u=\dfrac{1}{\sqrt{x^2+y^2+z^2}}$ , prove that $\dfrac{\partial^2u}{\partial x^2}+\dfrac{\partial^2u}{\partial y^2}+\dfrac{\partial^2u}{\partial z^2}=0$ .

Write $\rho=x^2+y^2+z^2$ , so $u=\rho^{-1/2}$ and $\dfrac{\partial\rho}{\partial x}=2x$ . Then

\frac{\partial u}{\partial x}=-\tfrac12\rho^{-3/2}\cdot2x=-x\,\rho^{-3/2}

\frac{\partial^2u}{\partial x^2}=-\rho^{-3/2}+x\cdot\tfrac32\rho^{-5/2}\cdot2x=3x^2\rho^{-5/2}-\rho^{-3/2}

By symmetry the $y$ and $z$ versions are $3y^2\rho^{-5/2}-\rho^{-3/2}$ and $3z^2\rho^{-5/2}-\rho^{-3/2}$ . Add all three:

\begin{aligned} u_{xx}+u_{yy}+u_{zz}&=3(x^2+y^2+z^2)\rho^{-5/2}-3\rho^{-3/2}\\ &=3\rho\cdot\rho^{-5/2}-3\rho^{-3/2}=3\rho^{-3/2}-3\rho^{-3/2}=\mathbf{0}\ \checkmark \end{aligned}

Example 6. If $u=x^2\tan^{-1}\!\dfrac yx-y^2\tan^{-1}\!\dfrac xy$ , show that $\dfrac{\partial^2u}{\partial x\,\partial y}=\dfrac{x^2-y^2}{x^2+y^2}$ .

First differentiate with respect to $y$ . For the first term:

x^2\cdot\frac{1}{1+y^2/x^2}\cdot\frac1x=\frac{x^3}{x^2+y^2}

For the second term (product rule):

\frac{\partial}{\partial y}\left[y^2\tan^{-1}\frac xy\right]=2y\tan^{-1}\frac xy+y^2\cdot\frac{1}{1+x^2/y^2}\cdot\left(-\frac{x}{y^2}\right)=2y\tan^{-1}\frac xy-\frac{xy^2}{x^2+y^2}

Subtract the second from the first, and note that $x^3+xy^2=x(x^2+y^2)$ :

\frac{\partial u}{\partial y}=\frac{x^3+xy^2}{x^2+y^2}-2y\tan^{-1}\frac xy=x-2y\tan^{-1}\frac xy

Now differentiate with respect to $x$ :

\frac{\partial^2u}{\partial x\,\partial y}=1-2y\cdot\frac{1}{1+x^2/y^2}\cdot\frac1y=1-\frac{2y^2}{x^2+y^2}=\frac{x^2+y^2-2y^2}{x^2+y^2}=\mathbf{\frac{x^2-y^2}{x^2+y^2}}\ \checkmark

Study tip. Simplify after the first derivative, before differentiating again. In Example 6 the whole expression becomes $x-2y\tan^{-1}(x/y)$ , which makes the second step easy.

Watch out. For $f_{xx}$ you differentiate with respect to $x$ again and still treat $y$ as a constant. Also, do not confuse $f_{xx}$ with $(f_x)^2$ . They are different things.

Page 8 of 13

Chain rule, part 1: one parameter

The idea

If $z=f(x,y)$ and both $x$ and $y$ depend on one variable $t$ , then $z$ becomes a function of $t$ alone. The chain rule gives $\dfrac{dz}{dt}$ without substituting first.

Chain rule (one parameter). If $z=f(x,y)$ , $x=x(t)$ , $y=y(t)$ , then

$\frac{dz}{dt}=\frac{\partial z}{\partial x}\cdot\frac{dx}{dt}+\frac{\partial z}{\partial y}\cdot\frac{dy}{dt}$

The tree diagram

Draw the dependencies as a tree: $z$ at the top, the variables it depends on below it, and the parameter at the bottom.

graph TD
    z -->|∂z/∂x| x
    z -->|∂z/∂y| y
    x -->|dx/dt| t
    y -->|dy/dt| t

How to read the tree. For each path from $z$ down to $t$ , take the product of the derivatives along its branches. Then add all the paths. Here there are two paths ( $z\to x\to t$ and $z\to y\to t$ ), so there are two terms.

With three variables $w=f(x,y,z)$ there are three paths:

\frac{dw}{dt}=\frac{\partial w}{\partial x}\frac{dx}{dt}+\frac{\partial w}{\partial y}\frac{dy}{dt}+\frac{\partial w}{\partial z}\frac{dz}{dt}

Method

Draw the tree.
Find the partial derivatives of the outer function with respect to each middle variable.
Find the derivative of each middle variable with respect to $t$ .
Multiply along each path, and then add.
Put $x(t)$ , $y(t)$ , $\dots$ so that the answer is in terms of $t$ (or put the given value of $t$ ).

Examples

Example 1. $z=5x^3y^2$ , $x=t^3$ , $y=t^5$ . Find $\dfrac{dz}{dt}$ , and check by substituting first.

\frac{\partial z}{\partial x}=15x^2y^2\quad\frac{\partial z}{\partial y}=10x^3y\quad\frac{dx}{dt}=3t^2\quad\frac{dy}{dt}=5t^4

\begin{aligned} \frac{dz}{dt}&=15x^2y^2\cdot3t^2+10x^3y\cdot5t^4\\ &=45(t^3)^2(t^5)^2\,t^2+50(t^3)^3\,t^5\,t^4\\ &=45t^{18}+50t^{18}=\mathbf{95t^{18}} \end{aligned}

Check. If we substitute first, $z=5(t^3)^3(t^5)^2=5t^{19}$ , so $\dfrac{dz}{dt}=95t^{18}$ . $\checkmark$

Example 2. $z=\log(3x^2+y)$ , $x=\sqrt t$ , $y=t^{1/3}$ . Find $\dfrac{dz}{dt}$ .

\frac{\partial z}{\partial x}=\frac{6x}{3x^2+y}\quad\frac{\partial z}{\partial y}=\frac{1}{3x^2+y}\quad\frac{dx}{dt}=\frac{1}{2\sqrt t}\quad\frac{dy}{dt}=\frac13t^{-2/3}

With $x=\sqrt t$ we get $3x^2+y=3t+t^{1/3}$ , so

\frac{dz}{dt}=\frac{6\sqrt t}{3t+t^{1/3}}\cdot\frac{1}{2\sqrt t}+\frac{1}{3t+t^{1/3}}\cdot\frac13t^{-2/3} =\frac{3+\tfrac13t^{-2/3}}{3t+t^{1/3}}

Multiply the top and the bottom by $3t^{2/3}$ to simplify:

\frac{dz}{dt}=\mathbf{\frac{9t^{2/3}+1}{3t\left(3t^{2/3}+1\right)}}

Example 3. Find $\dfrac{dw}{dt}$ for $w=x^2+y^2$ along the path $x=at^2$ , $y=2at$ .

\frac{\partial w}{\partial x}=2x\quad\frac{\partial w}{\partial y}=2y\quad\frac{dx}{dt}=2at\quad\frac{dy}{dt}=2a

\frac{dw}{dt}=2(at^2)(2at)+2(2at)(2a)=\mathbf{4a^2t^3+8a^2t}

(The path is a parabola, and $w$ is the square of the distance from the origin along it.)

Example 4. Find $\dfrac{dw}{dt}$ at $t=3$ if $w=\ln(x^2+y^2+z^2)$ , $x=\cos t$ , $y=\sin t$ , $z=4\sqrt t$ .

\frac{\partial w}{\partial x}=\frac{2x}{x^2+y^2+z^2}\quad\text{and similarly for }y,z;\qquad \frac{dx}{dt}=-\sin t\quad\frac{dy}{dt}=\cos t\quad\frac{dz}{dt}=\frac{2}{\sqrt t}

Here $x^2+y^2+z^2=\cos^2t+\sin^2t+16t=1+16t$ , so:

\begin{aligned} \frac{dw}{dt}&=\frac{2\cos t(-\sin t)+2\sin t\cos t+2(4\sqrt t)\cdot\frac{2}{\sqrt t}}{1+16t}\\[4pt] &=\frac{-2\sin t\cos t+2\sin t\cos t+16}{1+16t}=\frac{16}{1+16t} \end{aligned}

At $t=3$ :

\left.\frac{dw}{dt}\right|_{t=3}=\frac{16}{1+48}=\mathbf{\frac{16}{49}}

Watch out. Use $\partial$ for the outer function (it has many variables) and $d$ for the inner ones (each has only $t$ ). Do not forget the second path. Stopping after the $x$ term is the most common mistake.

Page 9 of 13

Chain rule, part 2: two parameters

The idea

Now let the middle variables depend on two parameters. If $z=f(x,y)$ with $x=x(u,v)$ and $y=y(u,v)$ , then $z$ is a function of $u$ and $v$ , and we need partial derivatives with respect to each.

Chain rule (two parameters).

$\frac{\partial z}{\partial u}=\frac{\partial z}{\partial x}\cdot\frac{\partial x}{\partial u}+\frac{\partial z}{\partial y}\cdot\frac{\partial y}{\partial u}$

$\frac{\partial z}{\partial v}=\frac{\partial z}{\partial x}\cdot\frac{\partial x}{\partial v}+\frac{\partial z}{\partial y}\cdot\frac{\partial y}{\partial v}$

flowchart TD
    Z[z] --- X[x]
    Z --- Y[y]
    X --- U1[u]
    X --- V1[v]
    Y --- U2[u]
    Y --- V2[v]

The tree rule is the same. For $\dfrac{\partial z}{\partial u}$ : take every path from $z$ down to $u$ , multiply along each path, and add. For $\dfrac{\partial z}{\partial v}$ : take every path down to $v$ .

Examples

Example 1. $z=x^2-y\tan x$ , $x=\dfrac uv$ , $y=uv$ . Find $\dfrac{\partial z}{\partial u}$ and $\dfrac{\partial z}{\partial v}$ .

The pieces:

\frac{\partial z}{\partial x}=2x-y\sec^2x\quad\frac{\partial z}{\partial y}=-\tan x

\frac{\partial x}{\partial u}=\frac1v\quad\frac{\partial x}{\partial v}=-\frac{u}{v^2}\quad\frac{\partial y}{\partial u}=v\quad\frac{\partial y}{\partial v}=u

Combine:

\frac{\partial z}{\partial u}=(2x-y\sec^2x)\cdot\frac1v+(-\tan x)\cdot v =\frac{2x-y\sec^2x}{v}-v\tan x

\frac{\partial z}{\partial v}=(2x-y\sec^2x)\cdot\left(-\frac{u}{v^2}\right)+(-\tan x)\cdot u

Finally, put $x=\dfrac uv$ and $y=uv$ to write the answers in terms of $u$ and $v$ :

\frac{\partial z}{\partial u}=\mathbf{\frac{2u}{v^2}-u\sec^2\frac uv-v\tan\frac uv}

\frac{\partial z}{\partial v}=\mathbf{-\frac{2u^2}{v^3}+\frac{u^2}{v}\sec^2\frac uv-u\tan\frac uv}

Example 2. $z=e^{x^2y}$ , $x=\sqrt{uv}$ , $y=\dfrac1v$ . Find $\dfrac{\partial z}{\partial u}$ and $\dfrac{\partial z}{\partial v}$ .

\frac{\partial z}{\partial x}=2xy\,e^{x^2y}\quad\frac{\partial z}{\partial y}=x^2e^{x^2y}

\frac{\partial x}{\partial u}=\frac{v}{2\sqrt{uv}}\quad\frac{\partial x}{\partial v}=\frac{u}{2\sqrt{uv}}\quad\frac{\partial y}{\partial u}=0\quad\frac{\partial y}{\partial v}=-\frac1{v^2}

Note that $x^2y=uv\cdot\dfrac1v=u$ , so every exponential is $e^{u}$ .

\frac{\partial z}{\partial u}=2xy\,e^{u}\cdot\frac{v}{2\sqrt{uv}}+0 =2\sqrt{uv}\cdot\frac1v\cdot\frac{v}{2\sqrt{uv}}\,e^{u}=\mathbf{e^{u}}

\frac{\partial z}{\partial v}=2xy\,e^{u}\cdot\frac{u}{2\sqrt{uv}}+x^2e^{u}\cdot\left(-\frac1{v^2}\right) =\frac uv\,e^{u}-\frac{uv}{v^2}\,e^{u}=\mathbf{0}

Check. If we substitute first, $z=e^{x^2y}=e^{u}$ depends on $u$ only, so $z_u=e^u$ and $z_v=0$ . $\checkmark$

Example 3. Express $\dfrac{\partial w}{\partial r}$ and $\dfrac{\partial w}{\partial s}$ in terms of $r$ and $s$ , if $w=x^2+y^2$ , $x=r-s$ , $y=r+s$ .

\frac{\partial w}{\partial x}=2x\quad\frac{\partial w}{\partial y}=2y\quad \frac{\partial x}{\partial r}=1\quad\frac{\partial y}{\partial r}=1\quad \frac{\partial x}{\partial s}=-1\quad\frac{\partial y}{\partial s}=1

\frac{\partial w}{\partial r}=2x\cdot1+2y\cdot1=2(r-s)+2(r+s)=\mathbf{4r}

\frac{\partial w}{\partial s}=2x\cdot(-1)+2y\cdot1=-2(r-s)+2(r+s)=\mathbf{4s}

Check. $w=(r-s)^2+(r+s)^2=2r^2+2s^2$ , so $w_r=4r$ and $w_s=4s$ . $\checkmark$

Example 4. Three middle variables. Find the first-order partial derivatives of $w=x+2y+z^2$ , where $x=\dfrac rs$ , $y=r^2+\ln s$ , $z=2r$ .

Here $z$ depends on $r$ only, so its branch goes to $r$ but not to $s$ .

flowchart TD
    W[w] --- X[x]
    W --- Y[y]
    W --- Z[z]
    X --- R1[r]
    X --- S1[s]
    Y --- R2[r]
    Y --- S2[s]
    Z --- R3[r]

\frac{\partial w}{\partial x}=1\quad\frac{\partial w}{\partial y}=2\quad\frac{\partial w}{\partial z}=2z

\frac{\partial x}{\partial r}=\frac1s\quad\frac{\partial y}{\partial r}=2r\quad\frac{\partial z}{\partial r}=2\qquad \frac{\partial x}{\partial s}=-\frac{r}{s^2}\quad\frac{\partial y}{\partial s}=\frac1s

Three paths down to $r$ , two paths down to $s$ :

\frac{\partial w}{\partial r}=1\cdot\frac1s+2\cdot2r+2z\cdot2=\frac1s+4r+8r=\mathbf{\frac1s+12r}

\frac{\partial w}{\partial s}=1\cdot\left(-\frac{r}{s^2}\right)+2\cdot\frac1s=\mathbf{-\frac{r}{s^2}+\frac2s}

When a variable has two roles

Sometimes the last variable also appears directly in the function. Then it gives both a direct path and the paths through the other variables.

Example 5. Find $\dfrac{dw}{dx}$ at $x=0$ for $w=xy+yz$ , with $y=\sin x$ and $z=e^x$ .

flowchart TD
    W[w] --- X1[x]
    W --- Y[y]
    W --- Z[z]
    Y --- X2[x]
    Z --- X3[x]

There are three paths from $w$ to $x$ : directly, through $y$ , and through $z$ .

\frac{dw}{dx}=\frac{\partial w}{\partial x}+\frac{\partial w}{\partial y}\cdot\frac{dy}{dx}+\frac{\partial w}{\partial z}\cdot\frac{dz}{dx}

With $\dfrac{\partial w}{\partial x}=y$ , $\dfrac{\partial w}{\partial y}=x+z$ , $\dfrac{\partial w}{\partial z}=y$ , $\dfrac{dy}{dx}=\cos x$ , $\dfrac{dz}{dx}=e^x$ :

\frac{dw}{dx}=y+(x+z)\cos x+y\,e^x

At $x=0$ we have $y=\sin0=0$ and $z=e^0=1$ , so

\left.\frac{dw}{dx}\right|_{x=0}=0+(0+1)(1)+0=\mathbf{1}

Check. If we substitute first, $w=x\sin x+e^x\sin x$ . Its derivative at $0$ is $0+0+1+0=1$ . $\checkmark$

Watch out. In Example 5, $\dfrac{\partial w}{\partial x}$ (treat $y$ and $z$ as constants) and $\dfrac{dw}{dx}$ (let them change with $x$ ) are different quantities. Also, the point $(0,1,2)$ in the original question does not lie on this curve. The curve passes through $(0,0,1)$ at $x=0$ .

Page 10 of 13

Related rates

The idea

If a quantity $z$ depends on $x$ and $y$ , and $x$ and $y$ are changing with time, the chain rule gives the rate of change of $z$ :

\frac{dz}{dt}=\frac{\partial z}{\partial x}\cdot\frac{dx}{dt}+\frac{\partial z}{\partial y}\cdot\frac{dy}{dt}

Method

Name the variables and write the formula $z=f(x,y)$ that connects them.
Write the given rates $\dfrac{dx}{dt}$ , $\dfrac{dy}{dt}$ . A quantity that is decreasing has a negative rate. A quantity that is constant has rate $0$ .
Find $\dfrac{\partial z}{\partial x}$ and $\dfrac{\partial z}{\partial y}$ .
Use the chain rule, and then put the values at that moment.
Write the answer with units, and say whether the quantity is increasing or decreasing.

Examples

Example 1. Area of a rectangle. The length and width of a rectangle are increasing at $1.5\ \text{cm/s}$ and $0.5\ \text{cm/s}$ . How fast is the area increasing when the length is $40\ \text{cm}$ and the width is $30\ \text{cm}$ ?

Let $x$ = length, $y$ = width, $A=xy$ . Then $\dfrac{dx}{dt}=1.5$ and $\dfrac{dy}{dt}=0.5$ , with $\dfrac{\partial A}{\partial x}=y$ and $\dfrac{\partial A}{\partial y}=x$ .

\frac{dA}{dt}=y\cdot1.5+x\cdot0.5=30(1.5)+40(0.5)=45+20=\mathbf{65\ \text{cm}^2/\text{s}}

The unit is square centimetre per second, because $A$ is an area.

Example 2. Volume of a cone. The height of a right circular cone is increasing at $0.2\ \text{cm/s}$ and the radius of its base is decreasing at $0.3\ \text{cm/s}$ . How fast is the volume changing when the height is $15\ \text{cm}$ and the radius is $10\ \text{cm}$ ?

Let $x$ = radius, $y$ = height, $V=\tfrac13\pi x^2y$ . The radius is decreasing, so $\dfrac{dx}{dt}=-0.3$ , and $\dfrac{dy}{dt}=0.2$ .

\frac{\partial V}{\partial x}=\frac{2\pi xy}{3}\qquad\frac{\partial V}{\partial y}=\frac{\pi x^2}{3}

\frac{dV}{dt}=\frac{2\pi xy}{3}(-0.3)+\frac{\pi x^2}{3}(0.2)

At $x=10$ , $y=15$ :

\frac{dV}{dt}=\frac{2\pi(150)}{3}(-0.3)+\frac{\pi(100)}{3}(0.2)=-30\pi+\frac{20\pi}{3}=\mathbf{-\frac{70\pi}{3}\ \text{cm}^3/\text{s}}\approx-73.3\ \text{cm}^3/\text{s}

The volume is decreasing: the decreasing radius (which is squared in the formula) has more effect than the increasing height.

Example 3. A quantity kept constant. Given $z=2xy^2-3x^2y$ and that $x$ is increasing at $2\ \text{cm/s}$ , find the rate at which $y$ changes when $x=3$ and $y=1$ , if $z$ stays constant.

Constant $z$ means $\dfrac{dz}{dt}=0$ . With $\dfrac{\partial z}{\partial x}=2y^2-6xy$ and $\dfrac{\partial z}{\partial y}=4xy-3x^2$ :

0=(2y^2-6xy)\cdot2+(4xy-3x^2)\frac{dy}{dt}

At $(3,1)$ : $\dfrac{\partial z}{\partial x}=2-18=-16$ and $\dfrac{\partial z}{\partial y}=12-27=-15$ . So

0=(-16)(2)+(-15)\frac{dy}{dt}\ \Rightarrow\ \frac{dy}{dt}=\mathbf{-\frac{32}{15}\ \text{cm/s}}

So $y$ is decreasing at about $2.13\ \text{cm/s}$ .

Note on the source. The handwritten page moved a bracket to the other side of the equals sign without changing its sign, and ended with $+\tfrac{32}{15}$ . When we put the numbers first (as above), we get $-\tfrac{32}{15}$ . The page also gave the rate of the cone's height as $0.2\ \text{m/s}$ , but it must be $0.2\ \text{cm/s}$ to match the other units. One middle line in the cone problem had $\tfrac{140\pi}{3}$ before reaching the correct $\tfrac{70\pi}{3}$ .

Watch out. Check signs and units at the end. A decreasing quantity needs a negative rate in the beginning. Also, all rates must be in the same units (here all in cm and seconds).

Page 11 of 13

Implicit differentiation

The idea

Sometimes $y$ is given implicitly by an equation such as $x^3+y^3=3axy$ , which cannot be solved easily for $y$ . Partial derivatives give a one-step formula for $\dfrac{dy}{dx}$ .

Implicit differentiation formula. If $f(x,y)=c$ defines $y$ as a function of $x$ , then

$\frac{dy}{dx}=-\frac{\partial f/\partial x}{\partial f/\partial y}=-\frac{f_x}{f_y}\qquad(f_y\neq0)$

Why it works

Along the curve $f(x,y)=c$ the value of $f$ does not change, so $\dfrac{d}{dx}f(x,y)=0$ . The chain rule (with $x$ having both roles, as on the previous page) gives

f_x\cdot1+f_y\cdot\frac{dy}{dx}=0

and solving for $\dfrac{dy}{dx}$ gives the formula.

Method

Take everything to one side so that the equation becomes $f(x,y)=0$ (or $=c$ ).
Find $f_x$ and $f_y$ as normal partial derivatives.
Write $\dfrac{dy}{dx}=-\dfrac{f_x}{f_y}$ . Keep the minus sign in front of the whole fraction.

Examples

Example 1. Using partial derivatives, find $\dfrac{dy}{dx}$ if $x^3+y^3=3axy$ . Then verify using ordinary implicit differentiation.

Let $f(x,y)=x^3+y^3-3axy$ . Then

f_x=3x^2-3ay\qquad f_y=3y^2-3ax

\frac{dy}{dx}=-\frac{3x^2-3ay}{3y^2-3ax}=\mathbf{\frac{ay-x^2}{y^2-ax}}

Verification. Differentiate both sides of $x^3+y^3=3axy$ with respect to $x$ :

3x^2+3y^2\frac{dy}{dx}=3a\left(x\frac{dy}{dx}+y\right)

Collect the $\dfrac{dy}{dx}$ terms:

\frac{dy}{dx}\left(3y^2-3ax\right)=3ay-3x^2\ \Rightarrow\ \frac{dy}{dx}=\frac{3ay-3x^2}{3y^2-3ax}=\frac{ay-x^2}{y^2-ax}

Both methods agree. $\checkmark$

Note on the source. The handwritten page wrote the first answer as $\dfrac{-3x^2-3ay}{3y^2-3ax}$ , where the minus sign was applied wrongly. It should be $-(3x^2-3ay)=-3x^2+3ay$ . The second method on the page gave the correct result.

Example 2. Find $\dfrac{dy}{dx}$ for the circle $x^2+y^2=a^2$ .

Let $f(x,y)=x^2+y^2-a^2$ , so $f_x=2x$ and $f_y=2y$ :

\frac{dy}{dx}=-\frac{2x}{2y}=\mathbf{-\frac xy}

This is the slope of the circle at $(x,y)$ . It agrees with geometry: the tangent to a circle is perpendicular to the radius, which has slope $y/x$ .

Watch out. Students make three mistakes here. (1) Missing the minus sign. (2) Writing $f_y$ on top and $f_x$ at the bottom. It is $f_x$ over $f_y$ . (3) Forgetting that the formula needs the equation in the form $f(x,y)=c$ , with all terms taken to one side first.

Page 12 of 13

Level curves

The idea

A level curve is the set of points where a function has one fixed value. Think of the contour lines on a map: each line joins the places that have the same height.

Definition. A level curve of $f(x,y)$ is the curve

$f(x,y)=k$

for a constant $k$ .

Level curves never cross (a point cannot have two different values of $f$ ). Where the curves are close together, the surface is steep. Where they are far apart, the surface is gentle.

Method

Set $f(x,y)=k$ for each requested value of $k$ .
Simplify each equation to a known curve: a line, circle, parabola, ellipse, and so on.
Draw all the curves together, and write the $k$ value on each.
To find the level curve through a given point, put the point in $f$ . That value is $k$ .

Examples

Example 1. Find the level curves of $f(x,y)=\sqrt{9-x^2-y^2}$ for $c=0,1,2,3$ .

Set $\sqrt{9-x^2-y^2}=c$ and square both sides:

9-x^2-y^2=c^2\ \Rightarrow\ x^2+y^2=9-c^2

$c$	Equation	Curve
$0$	$x^2+y^2=9$	circle of radius $3$
$1$	$x^2+y^2=8$	circle of radius $2\sqrt2\approx2.83$
$2$	$x^2+y^2=5$	circle of radius $\sqrt5\approx2.24$
$3$	$x^2+y^2=0$	the single point $(0,0)$

{
  "boundingBox": [-3.5, 3.5, 3.5, -3.5],
  "axis": true,
  "points": [
    { "id": "O", "x": 0, "y": 0, "label": "c = 3" }
  ],
  "segments": [],
  "circles": [
    { "center": "O", "radius": 3 },
    { "center": "O", "radius": 2.8284 },
    { "center": "O", "radius": 2.2361 }
  ],
  "functions": [],
  "title": "Level curves of f(x, y) = √(9 − x² − y²): from the outside in, c = 0, 1, 2; the point at the centre is c = 3"
}

The graph of $f$ is the upper half of a sphere of radius $3$ , and its level curves are the circles at each height. Notice that they come close together near the edge, where the surface is steepest.

Watch out. For $c>3$ there is no level curve, because $x^2+y^2=9-c^2$ would be negative. This agrees with the range $[0,3]$ found on the domain page. Also, $c$ must be $\ge0$ , because a square root cannot be negative.

Example 2. Find the level curves of $f(x,y)=x^2+2y$ for $k=-2,0,2$ .

Set $x^2+2y=k$ . Solving for $y$ gives a parabola:

y=\frac{k-x^2}{2}

$k$	Equation	Vertex
$-2$	$x^2=-2(y+1)$	$(0,-1)$
$0$	$x^2=-2y$	$(0,0)$
$2$	$x^2=-2(y-1)$	$(0,1)$

{
  "boundingBox": [-3, 2.5, 3, -3.5],
  "axis": true,
  "points": [],
  "segments": [],
  "circles": [],
  "parabolas": [
    { "vertex": [0, 1], "a": -0.5, "label": "k = 2" },
    { "vertex": [0, 0], "a": -0.5, "label": "k = 0" },
    { "vertex": [0, -1], "a": -0.5, "label": "k = −2" }
  ],
  "functions": [],
  "title": "Level curves of f(x, y) = x² + 2y: parabolas opening downward"
}

These are identical parabolas. Each one moves up by one unit when $k$ increases by $2$ .

Parabola shape. $x^2=4ay$ opens upward and $x^2=-4ay$ opens downward (here $4a=2$ , and the minus sign gives downward).

Example 3. Find and sketch the level curve of $f(x,y)=x^2+y^2$ that passes through $(2,0)$ .

Put the point to find the level:

k=f(2,0)=2^2+0^2=4

so the level curve is $x^2+y^2=4$ : a circle of radius $2$ centred at the origin.

{
  "boundingBox": [-3.5, 3.5, 3.5, -3.5],
  "axis": true,
  "points": [
    { "id": "O", "x": 0, "y": 0, "label": "" },
    { "id": "P1", "x": 2, "y": 0, "label": "(2,0)" },
    { "id": "P2", "x": 0, "y": 2, "label": "(0,2)" },
    { "id": "P3", "x": -2, "y": 0, "label": "(-2,0)" },
    { "id": "P4", "x": 0, "y": -2, "label": "(0,-2)" }
  ],
  "segments": [],
  "circles": [
    { "center": "O", "radius": 2 }
  ],
  "functions": [],
  "title": "Level curve x² + y² = 4: a circle of radius 2"
}

Page 13 of 13

Practice questions and formula sheet

Practice questions

Try these without looking at the answers. The answers are given below.

A. Limits and continuity

$\displaystyle\lim_{(x,y)\to(1,2)}\frac{x^2y}{x+y}$
Show that $\displaystyle\lim_{(x,y)\to(0,0)}\frac{x^2}{x^2+y^2}$ does not exist.
$\displaystyle\lim_{(x,y)\to(0,0)}\frac{x^2-y^2}{x-y}$
Let $f(x,y)=\dfrac{x^3}{x^2+y^2}$ for $(x,y)\ne(0,0)$ and $f(0,0)=0$ . Is $f$ continuous at the origin? (Hint: compare $|f|$ with $|x|$ .)

B. Partial derivatives

Find $f_x$ and $f_y$ for $f=x^3-3xy^2+e^{xy}$ .
Find $z_x$ and $z_y$ for $z=\dfrac{x}{x^2+y^2}$ .
Verify that $f_{xy}=f_{yx}$ for $f=x^2y^3+\sin(xy)$ .
Show that $z=\ln(x^2+y^2)$ satisfies $z_{xx}+z_{yy}=0$ .

C. Chain rule

Find $\dfrac{dz}{dt}$ for $z=x^2y$ , $x=\cos t$ , $y=\sin t$ .
Find $\dfrac{\partial w}{\partial u}$ and $\dfrac{\partial w}{\partial v}$ for $w=xy$ , $x=u+v$ , $y=u-v$ .

D. Implicit differentiation. Find $\dfrac{dy}{dx}$ :

$x^2+xy+y^2=7$
$\sin(xy)=x+y$

E. Rates, level curves, revision

The radius of a cylinder is increasing at $0.5\ \text{cm/s}$ and its height is decreasing at $1\ \text{cm/s}$ . How fast is the volume $V=\pi r^2h$ changing when $r=3\ \text{cm}$ and $h=10\ \text{cm}$ ?
Find the level curve of $f(x,y)=x^2+4y^2$ through $(2,1)$ and say what shape it is.
From Module 1. Find $\dfrac{d^2y}{dx^2}$ for $y=\dfrac{(3x^3+2x)(3x+5)}{x^4}$ .

Answers

$\dfrac{2}{3}$ . Put the values: $\dfrac{1\cdot2}{1+2}$ .
Along $y=mx$ the value is $\dfrac1{1+m^2}$ . It is $1$ for $m=0$ and $\tfrac12$ for $m=1$ . The values are different, so the limit does not exist.
$0$ . Factorise: $\dfrac{(x-y)(x+y)}{x-y}=x+y\to0$ .
Continuous. $\left|\dfrac{x^3}{x^2+y^2}\right|\le\dfrac{|x|^3}{x^2}=|x|\to0=f(0,0)$ , so by the sandwich theorem the limit equals $f(0,0)$ .
$f_x=3x^2-3y^2+y\,e^{xy}$ , $\quad f_y=-6xy+x\,e^{xy}$
$z_x=\dfrac{y^2-x^2}{(x^2+y^2)^2}$ , $\quad z_y=-\dfrac{2xy}{(x^2+y^2)^2}$
$f_x=2xy^3+y\cos(xy)$ and $f_y=3x^2y^2+x\cos(xy)$ . Both mixed derivatives equal $6xy^2+\cos(xy)-xy\sin(xy)$ .
$z_x=\dfrac{2x}{x^2+y^2}$ gives $z_{xx}=\dfrac{2y^2-2x^2}{(x^2+y^2)^2}$ . By symmetry $z_{yy}=\dfrac{2x^2-2y^2}{(x^2+y^2)^2}$ . The sum is $0$ .
$\dfrac{dz}{dt}=2xy(-\sin t)+x^2\cos t=\cos^3t-2\sin^2t\cos t$ . (Check: $z=\cos^2t\sin t$ .)
$w_u=y+x=2u$ , $\quad w_v=y-x=-2v$ . (Check: $w=u^2-v^2$ .)
$\dfrac{dy}{dx}=-\dfrac{2x+y}{x+2y}$
With $f=\sin(xy)-x-y$ : $\dfrac{dy}{dx}=-\dfrac{y\cos(xy)-1}{x\cos(xy)-1}=\dfrac{1-y\cos(xy)}{x\cos(xy)-1}$
$\dfrac{dV}{dt}=2\pi rh\cdot\dfrac{dr}{dt}+\pi r^2\cdot\dfrac{dh}{dt}=2\pi(30)(0.5)+\pi(9)(-1)=30\pi-9\pi=21\pi\approx66.0\ \text{cm}^3/\text{s}$ (increasing).
$k=4+4=8$ . The level curve is $x^2+4y^2=8$ , that is $\dfrac{x^2}{8}+\dfrac{y^2}{2}=1$ : an ellipse with semi-axes $2\sqrt2$ (along $x$ ) and $\sqrt2$ (along $y$ ).
Simplify first: $y=9+\dfrac{15}{x}+\dfrac{6}{x^2}+\dfrac{10}{x^3}$ . Then $y'=-\dfrac{15}{x^2}-\dfrac{12}{x^3}-\dfrac{30}{x^4}$ and $y''=\dfrac{30}{x^3}+\dfrac{36}{x^4}+\dfrac{120}{x^5}=\dfrac{30x^2+36x+120}{x^5}$ .

Study tip. In 15, if you expand and divide before differentiating, you can use the simple power rule instead of a long quotient rule. Use this shortcut whenever the denominator has only one term.

Formula sheet

Functions, limits and continuity


Domain	no zero denominators, no even roots of negative numbers, no logs of zero or negative numbers
Limit	$\lim_{(x,y)\to(a,b)}f=L$ along every path
Substitution	works for polynomials, and for rational functions with non-zero denominator
$0/0$	factorise and cancel, or use $x-y=(\sqrt x-\sqrt y)(\sqrt x+\sqrt y)$
Does not exist	two paths give different values; try $y=mx$
Continuous at $(a,b)$	$f(a,b)$ defined, limit exists, limit $=f(a,b)$

Partial derivatives

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


$f_x$	differentiate in $x$ , treat $y$ as a constant
$f_y$	differentiate in $y$ , treat $x$ as a constant
Slope of surface	$f_x$ in the $x$ -direction, $f_y$ in the $y$ -direction
Second order	$f_{xx}$ , $f_{yy}$ , and mixed $f_{xy}=\dfrac{\partial^2f}{\partial y\,\partial x}$ , $f_{yx}=\dfrac{\partial^2f}{\partial x\,\partial y}$
Mixed partials	$f_{xy}=f_{yx}$ when they are continuous
Laplace's equation	$u_{xx}+u_{yy}+u_{zz}=0$

Chain rule

Situation	Formula
$z=f(x,y)$ , $x(t)$ , $y(t)$	$\dfrac{dz}{dt}=z_x\dfrac{dx}{dt}+z_y\dfrac{dy}{dt}$
$z=f(x,y)$ , $x(u,v)$ , $y(u,v)$	$\dfrac{\partial z}{\partial u}=z_x\dfrac{\partial x}{\partial u}+z_y\dfrac{\partial y}{\partial u}$ , and the same with $v$
$w=f(x,y,z)$ , all depend on $t$	$\dfrac{dw}{dt}=w_x\dfrac{dx}{dt}+w_y\dfrac{dy}{dt}+w_z\dfrac{dz}{dt}$
Variable in two roles	$\dfrac{dw}{dx}=\dfrac{\partial w}{\partial x}+\dfrac{\partial w}{\partial y}\dfrac{dy}{dx}+\dfrac{\partial w}{\partial z}\dfrac{dz}{dx}$
Tree rule	multiply along each path, add the paths

Implicit differentiation and level curves


Implicit	$f(x,y)=c\ \Rightarrow\ \dfrac{dy}{dx}=-\dfrac{f_x}{f_y}$
Level curve	$f(x,y)=k$
Through a point $(a,b)$	take $k=f(a,b)$
Related rates	quantity decreasing: negative rate. Held constant: rate $0$

Module 2 at a glance

What you will be able to do

How to use these notes

Where each topic sits

Functions of two variables

The idea

Domain and range

Limits of functions of two variables

The idea

Easy case: just substitute

When substitution gives 0/00/00/0

Showing a limit does not exist: the two-path test

Continuity

The idea

What you can take for granted

Method

Examples

Partial derivatives

The idea

Method

Examples

Three or more variables

Slope and rate of change

The idea

Examples

Second-order partial derivatives

The idea

Examples

Third order

Proving an equation with derivatives

Chain rule, part 1: one parameter

The idea

The tree diagram

Method

Examples

Chain rule, part 2: two parameters

The idea

Examples

When a variable has two roles

Related rates

The idea

Method

Examples

Implicit differentiation

The idea

Why it works

Method

Examples

Level curves

The idea

Method

Examples

Practice questions and formula sheet

Practice questions

Answers

Formula sheet

Functions, limits and continuity

Partial derivatives

Chain rule

Implicit differentiation and level curves

When substitution gives $0/0$