Please, please can you label the axes three dimensional graphs completely and more clearly? Your site is the best! I am 75 and trying to revise my university math of some 50 years ago! Many thanks

The up axis is always z, and you can use the right hand rule to see which the other two are; your index finger will be pointing at the x-axis, and you're middle finger will be pointing at the y-axis. (Note: different software use different conventions for labeling the axes. Sometimes z represents depth and y represents height. But mac grapher - the software Grant is using here - always uses the convention I described above.)

Why there is no a mission in multivariable funtions?

It's more about learning the techniques, notation, and order of operations than plugging and cranking a bunch of examples. It is also an area of math that isn't really intended to learn purely online. This section is more of a nudge to help learn or better understand something you are currently learning in a 5th semester math course. Usually proceeded by MA 161,162 and MA261,262 or something similar. I'm just using Purdues course #s.

The explanation with the graph is hard to follow. I see no relation between the graph being shown and the equations being listed. I don't understand any of it.

If you think back to Calculus 1 (or single-variable calculus), recall the the derivative of a function is equal to its slope at any point. If you don't understand that concept, it might be good to look back and review the section on derivatives. In this case, when we take a slice of the graph, the two-dimensional intersection of the graph and the plane looks like a single-variable function. Which variable the function is of depends on the orientation of the graph. If the graph is parallel to the x-axis, it looks like a function of x, and if the graph is parallel to the y-axis, the intersection looks like a function of y. The partial derivative is a way to find the slope in either the x or y direction, at the point indicated. By treating the other variable like a constant, the situation seems to simplify to something we can understand in terms of single-variable derivatives, which we learned in Calc 1. If you still do not understand, let me know, and we can try to work it out. Make sure you have learned previous lessons well.

Just a nitpicky question about notation, is the "∂" symbol also called del like this article says? I thought that "∇" was del, and that the partial symbol was just a stylized "d"

Per the Greek alphabet, ∂ is lowercase delta whereas ∇ is uppercase delta.

2 quick questions: 1) example 2, sin( 0 + pi) = 0... however I ran my calculator and got the answer .0548. I checked and my settings are in degrees, not radians. 2) In both my homework (Dynamic Atmosphere) and your examples, when deriving 'y' for example, why do we make '2x' or 'x^2' equal to zero. Why do we not bring them as a constant?

If they are stand alone such as x^2+2x+2xy^3. If you are taking the partial derivative with respect to y, you treat the others as a constant. The derivative of a constant is 0, so it becomes 0+0+2x(3y^2). You'll notice since the last one is multiplied by Y, you treat it as a constant multiplied by the derivative of the function. I.E d/dx 2x^2 is just 2(d/dx(x^2)) or 2(2x) or 4x. Hope this helps.

Main content

Course: Multivariable calculus > Unit 2

Lesson 3: Partial derivative and gradient (articles)

Introduction to partial derivatives

Google Classroom

What is the partial derivative, how do you compute it, and what does it mean?

Background

What we're building to

For a multivariable function, like $f (x, y) = x^{2} y$ ‍, computing partial derivatives looks something like this:

\begin{aligned} \frac{\partial f}{\partial x} & = \underset{\begin{array}{c} Treat y as constant; \\ take derivative. \end{array}}{\underset{⏟}{\frac{\partial}{\partial x} x^{2} y}} = 2 x y \\ \frac{\partial f}{\partial y} & = \underset{\begin{array}{c} Treat x as constant; \\ take derivative. \end{array}}{\underset{⏟}{\frac{\partial}{\partial y} x^{2} y}} = x^{2} \cdot 1 \end{aligned}

This swirly-d symbol, $\partial$ ‍, often called "del", is used to distinguish partial derivatives from ordinary single-variable derivatives. Or, should I say ... to differentiate them.
The reason for a new type of derivative is that when the input of a function is made up of multiple variables, we want to see how the function changes as we let just one of those variables change while holding all the others constant.
With respect to three-dimensional graphs, you can picture the partial derivative $\frac{\partial f}{\partial x}$ ‍ by slicing the graph of $f$ ‍ with a plane representing a constant $y$ ‍-value and measuring the slope of the resulting curve along the cut.

What is a partial derivative?

We'll assume you are familiar with the ordinary derivative

\frac{d f}{d x}

from single variable calculus. I actually quite like this notation for the derivative, because you can interpret it as follows:

Interpret $d x$ ‍ as "a very tiny change in $x$ ‍".
Interpret $d f$ ‍ as "a very tiny change in the output of $f$ ‍", where it is understood that this tiny change is whatever results from the tiny change $d x$ ‍ to the input.

In fact, I think this intuitive feel for the symbol

\frac{d f}{d x}

is one of the most useful takeaways from single-variable calculus, and when you really start feeling it in your bones, most of the concepts around derivatives start to click.

For example, when you apply it to the graph of

f

, you can interpret this "ratio"

\frac{d f}{d x}

as the rise-over-run slope of the graph of

f

, which depends on the point where you started.

How does this work for multivariable functions?

Consider some function with a two-dimensional input and a one-dimensional output.

f (x, y) = x^{2} - 2 x y

There's nothing stopping us from writing the same expression,

\frac{d f}{d x}

, and interpreting it the same way:

$d x$ ‍ can still represent a tiny change in the variable $x$ ‍, which is now just one component of our input.
$d f$ ‍ can still represent the resulting change to the output of the function $f (x, y)$ ‍.

However, this ignores the fact that there is another input variable

y

. The input space now has multiple dimensions, so we can change the input in many directions other than the

x

-direction. For example, what about changing

y

slightly by some small value

d y

? Now if we re-interpret

d f

to represent the tiny change to the function that this

d y

shift brings about, we would have a different derivative

\frac{d f}{d y}

Neither one of these derivatives tells the full story of how our function

f (x, y)

changes when its input changes slightly, so we call them partial derivatives. To emphasize the difference, we no longer use the letter $d$ ‍ to indicate tiny changes, but instead introduce a newfangled symbol $\partial$ ‍ to do the trick, writing each partial derivative as

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

\frac{\partial f}{\partial y}

, etc.

You read the symbol

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

out loud by saying "the partial derivative of

f

with respect to

x

\begin{matrix} Can be thought of as \\ "a tiny change in \\ the function’s output" \\ \begin{matrix} \begin{matrix} Used instead of "d" in \\ usual \frac{d f}{d x} notation to \\ emphasize that this is \\ a partial derivative. \end{matrix} & \begin{matrix} ↗^{\overset{⏞}{\partial f}} ↖ \\ — \\ ↘_{\underset{⏟}{\partial x}} ↙ \end{matrix} & \begin{matrix} Multivariable \\ function \\ Indicates which \\ input variable \\ is changed slightly. \end{matrix} \end{matrix} \\ Can be thought of as \\ "a tiny change in x " \end{matrix}

Example: Computing a partial derivative

Consider this function:

f (x, y) = x^{2} y^{3}

Suppose I asked you to evaluate

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

, the partial derivative with respect to

x

, at the input

(3, 2)

"What? But I haven't learned how yet!"

Don't worry, it's mostly just the same mechanics as an ordinary derivative.

From the introduction above, you should know that this is asking about the rate at which the output of

f

changes as we nudge the

x

-component of the input slightly, perhaps moving from

(3, 2)

(3.01, 2)

Since we only care about movement in the

x

-direction, we might as well treat the

y

-value as a constant. In fact, we can just plug in

y = 2

ahead of time before computing any derivatives:

f (x, 2) = x^{2} (2)^{3} = 8 x^{2}

Now, asking how

f

changes in response to a small shift in

x

is just an ordinary, single-variable derivative.

Concept check

What is the derivative of this function $f (x, 2) = 8 x^{2}$ ‍ evaluated at $x = 3$ ‍?

Without pre-evaluating $y$ ‍

Now suppose I asked you to find

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

, but I didn't ask you to evaluate it at a specific point. In other words, you should give me new multivariable function which takes any point

(x, y)

as its input and tells me what the rate of change of

f

near that point is as we move purely in the

x

-direction.

You can start the same way, treating the

y

value as a constant. However, this time, you cannot plug in an actual constant value, like

y = 2

. Instead, pretend that

y

is constant and take the derivative:

\frac{d}{d x} f (x, y) = \underset{Pretend y is constant}{\underset{⏟}{\frac{d}{d x} (x^{2} y^{3})}} = 2 x y^{3}

Or rather, since to emphasize that this is a multivariable function, we use the symbol

\partial

instead of

d

\frac{\partial}{\partial x} f (x, y) = \frac{\partial}{\partial x} (x^{2} y^{3}) = 2 x y^{3}

As a sanity check, you can plug in

(3, 2)

to see that we get the same result as above.

"So, what's the difference between

\frac{d}{d x}

and

\frac{\partial}{\partial x}

? They seem to be used the same way."

Honestly, as far as I'm concerned, there's not really a difference between these operations. You could be pedantic and say one is only defined for single variable functions. But as far as intuition and computation go, they are one and the same, and the difference is just meant to clarify what type of function is being differentiated.

Interpreting partial derivatives with graphs

Consider this function:

f (x, y) = \frac{1}{5} (x^{2} - 2 x y) + 3

Here is a video showing its graph rotating, just to get a feel for the three-dimensional nature of it.

Khan Academy video wrapper

See video transcript

Think about the partial derivative of

f

with respect to

x

, perhaps evaluated at the point

(2, 0)

\frac{\partial f}{\partial x} (2, 0)

In terms of the graph, what does the value of this expression tell us about the behavior of the function

f

at the point

(2, 0)

Treat $y$ ‍ as constant $\to$ ‍ slice graph with plane

The first step when computing this value is to treat

y

as a constant. Specifically, if we are limiting our view to what happens at the point

(2, 0)

, we should only look at the set of points where

y = 0

. In three-dimensional space, this set is plane perpendicular to the

y

-axis, passing through the origin.

This plane

y = 0

, shown in white, slices into the graph of

f (x, y)

along a parabolic curve, shown faintly in red. We can interpret

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

as giving the slope of a tangent line to this curve. Why? Because

\partial x

is a slight nudge in the

x

-direction, the run, and

\partial f

is the resulting change in the

z

-direction, the rise.

What about

\frac{\partial f}{\partial y}

at that same point

(2, 0)

? The points where

x = 2

also make up a plane, but this time it's a plane perpendicular to the

x

-axis intersecting the point

x = 2

. This slices the graph along a new curve, and

\frac{\partial f}{\partial y}

will give the slope of that new curve.

Reflection Question

In the picture to the right, the "curve" where the graph of

f (x, y) = \frac{1}{5} (x^{2} - 2 x y) + 3

intersects the plane defined by

x = 2

looks like it might be a straight line.

Is it really a line?

The partial derivative

\frac{\partial f}{\partial y}

evaluated at any point

(2, y)

gives the slope of this line,

- \frac{4}{5}

, no matter what

y

is. In fact, the partial derivative of this function doesn't depend on

y

at all:

\begin{aligned} \frac{\partial f}{\partial y} & = \frac{\partial}{\partial y} (\frac{1}{5} (x^{2} - 2 x y)) \\ = \frac{1}{5} (0 - 2 x) \\ = - \frac{2}{5} x \end{aligned}

Graphically, this means as we choose different values of

x

, and the plane representing the constant

x

value shifts left or right, it will always intersect the graph along a line, since its slope is constant with respect to

y

. Moreover, the slope of that line will always be

- \frac{2}{5}

times the value of

x

Khan Academy video wrapper

See video transcript

Phrasing and notation

Here are some of the phrases you might hear in reference to this

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

operation:

"The partial derivative of $f$ ‍ with respect to $x$ ‍"
"Del f, del x"
"Partial f, partial x"
"The partial derivative (of $f$ ‍) in the $x$ ‍-direction"

Alternate notation

In the same way that people sometimes prefer to write

f^{'}

instead of

\frac{d f}{d x}

, we have the following notation:

\begin{aligned} f_{x} & \leftrightarrow \frac{\partial f}{\partial x} \\ f_{y} & \leftrightarrow \frac{\partial f}{\partial y} \\ f_{⟨ Some variable ⟩} & \leftrightarrow \frac{\partial f}{\partial ⟨ That same variable ⟩} \end{aligned}

A note about "del"

While it's common to refer to the partial symbol

\partial

as "del", this can be confusing because "del" is also the name of the Nabla symbol

\nabla

, which we will introduce in the next article.

A more formal definition

Although thinking of

d x

\partial x

as really tiny changes in the value of

x

is a useful intuition, it is healthy to occasionally step back and remember that defining things precisely requires introducing limits. After all, what specific small value would

\partial x

be? One one hundredth? One one millionth?

10^{- 10^{10}}

The point of calculus is that we don't use any one tiny number, but instead consider all possible values and analyze what tends to happen as they approach a limiting value. The single variable derivative, for example, is defined like this:

\begin{array}{r} \frac{d f}{d x} (x_{0}) = lim_{h \to 0} \frac{f (x_{0} + h) - f (x_{0})}{h} \end{array}

$h$ ‍ represents the "tiny value" that we intuitively think of as $d x$ ‍.
The $h \to 0$ ‍ under the limit indicates that we care about very small values of $h$ ‍, those approaching $0$ ‍.
$f (x_{0} + h) - f (x_{0})$ ‍ is the change in the output that results from adding $h$ ‍ to the input, which is what we think of as $d f$ ‍.

Formally defining the partial derivative looks almost identical. If

f (x, y, \dots)

is a function with multiple inputs, here's how that looks:

\begin{aligned} \frac{\partial f}{\partial x} (x_{0}, y_{0}, \dots) & = lim_{h \to 0} \frac{f (x_{0} + h, y_{0}, \dots) - f (x_{0}, y_{0}, \dots)}{h} \end{aligned}

Similarly, here's how the partial derivative with respect to

y

looks:

\begin{aligned} \frac{\partial f}{\partial y} (x_{0}, y_{0}, \dots) & = lim_{h \to 0} \frac{f (x_{0}, y_{0} + h, \dots) - f (x_{0}, y_{0}, \dots)}{h} \end{aligned}

The point is that

h

, which represents a tiny tweak to the input, is added to different input variables depending on which partial derivative we are taking.

People will often refer to this as the limit definition of a partial derivative.

Reflection question: How can we think about this limit definition in the context of the graphical interpretation above? What is

h

? What does it look like for

h \to 0

Summary

For a multivariable function, like $f (x, y) = x^{2} y$ ‍, computing partial derivatives looks something like this:

\begin{aligned} \frac{\partial f}{\partial x} & = \underset{\begin{array}{c} Treat y as constant; \\ take derivative. \end{array}}{\underset{⏟}{\frac{\partial}{\partial x} x^{2} y}} = 2 x y \\ \frac{\partial f}{\partial y} & = \underset{\begin{array}{c} Treat x as constant; \\ take derivative. \end{array}}{\underset{⏟}{\frac{\partial}{\partial y} x^{2} y}} = x^{2} \cdot 1 \end{aligned}

This swirly-d symbol $\partial$ ‍, often called "del", is used to distinguish partial derivatives from ordinary single-variable derivatives.
The reason for a new type of derivative is that when the input of a function is made up of multiple variables, we want to see how the function changes as we let just one of those variables change while holding all the others constant.
With respect to three-dimensional graphs, you can picture the partial derivative $\frac{\partial f}{\partial x}$ ‍ by slicing the graph of $f$ ‍ with a plane representing a constant $y$ ‍-value, and measuring the slope of the resulting cut.

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Chopin1809
Posted 7 years ago. Direct link to Chopin1809's post “Please, please can you la...”
Please, please can you label the axes three dimensional graphs completely and
more clearly? Your site is the best! I am 75 and trying to revise my university
math of some 50 years ago!
Many thanks
Button navigates to signup pageComment on Chopin1809's post “Please, please can you la...”
(108 votes)
Answer
- Alia
  Posted a year ago. Direct link to Alia's post “The up axis is always z, ...”
  The up axis is always z, and you can use the right hand rule to see which the other two are; your index finger will be pointing at the x-axis, and you're middle finger will be pointing at the y-axis. (Note: different software use different conventions for labeling the axes. Sometimes z represents depth and y represents height. But mac grapher - the software Grant is using here - always uses the convention I described above.)
  Button navigates to signup page
  (3 votes)
Marcel Brown
Posted 8 years ago. Direct link to Marcel Brown's post “In Example 2, isn't cos(0...”
In Example 2, isn't cos(0*pi) = cos(0) = 1? And therefore the answer is non-negative?
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(22 votes)
Answer
DM
Posted 8 years ago. Direct link to DM's post “Where it says: the set of...”
Where it says: the set of all points where y = 0 includes all points of the form (x,0,z). I believe this set is mistakenly called the xy-plane. It should be called the xz-plane.
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(8 votes)
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oleksandr.grygorchuk
Posted 5 years ago. Direct link to oleksandr.grygorchuk's post “Hi, I find these articles...”
Hi,
I find these articles very useful, and I'd like to keep them at hand for future references. Are they available for download, in pdf or some other print-friendly format, somewhere? Thanks.
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(4 votes)
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l wx
Posted 8 years ago. Direct link to l wx's post “Why there is no a mission...”
Why there is no a mission in multivariable funtions?
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(1 vote)
Answer
- Chuck Batcheller
  Posted 8 years ago. Direct link to Chuck Batcheller's post “It's more about learning ...”
  It's more about learning the techniques, notation, and order of operations than plugging and cranking a bunch of examples. It is also an area of math that isn't really intended to learn purely online. This section is more of a nudge to help learn or better understand something you are currently learning in a 5th semester math course. Usually proceeded by MA 161,162 and MA261,262 or something similar. I'm just using Purdues course #s.
  Comment on Chuck Batcheller's post “It's more about learning ...”
  (5 votes)
GSrayes
Posted 8 years ago. Direct link to GSrayes's post “I still see the gradient ...”
I still see the gradient vary on the video graphical representation. Does not make sense.
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(1 vote)
Answer
Robert891201
Posted 3 years ago. Direct link to Robert891201's post “The explanation with the ...”
The explanation with the graph is hard to follow. I see no relation between the graph being shown and the equations being listed. I don't understand any of it.
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- Claire
  Posted 3 years ago. Direct link to Claire's post “If you think back to Calc...”
  If you think back to Calculus 1 (or single-variable calculus), recall the the derivative of a function is equal to its slope at any point. If you don't understand that concept, it might be good to look back and review the section on derivatives. In this case, when we take a slice of the graph, the two-dimensional intersection of the graph and the plane looks like a single-variable function. Which variable the function is of depends on the orientation of the graph. If the graph is parallel to the x-axis, it looks like a function of x, and if the graph is parallel to the y-axis, the intersection looks like a function of y. The partial derivative is a way to find the slope in either the x or y direction, at the point indicated. By treating the other variable like a constant, the situation seems to simplify to something we can understand in terms of single-variable derivatives, which we learned in Calc 1. If you still do not understand, let me know, and we can try to work it out. Make sure you have learned previous lessons well.
  Comment on Claire's post “If you think back to Calc...”
  (3 votes)
joonwonsuh
Posted 6 years ago. Direct link to joonwonsuh's post “when was this published?”
when was this published?
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- Davide Ghazal
  Posted 5 years ago. Direct link to Davide Ghazal's post “They kicked off with the ...”
  They kicked off with the playlist on multivariable calculus 4 or 5 years ago. Not sure about this current article though.
  Button navigates to signup page
  (1 vote)
Therese
Posted 6 years ago. Direct link to Therese's post “Just a nitpicky question ...”
Just a nitpicky question about notation, is the "∂" symbol also called del like this article says? I thought that "∇" was del, and that the partial symbol was just a stylized "d"
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- Mason Moreland
  Posted 5 years ago. Direct link to Mason Moreland's post “Per the Greek alphabet, ∂...”
  Per the Greek alphabet, ∂ is lowercase delta whereas ∇ is uppercase delta.
  Comment on Mason Moreland's post “Per the Greek alphabet, ∂...”
  (2 votes)
Brian Conlan
Posted 8 years ago. Direct link to Brian Conlan's post “2 quick questions: 1) exa...”
2 quick questions:
1) example 2, sin( 0 + pi) = 0... however I ran my calculator and got the answer .0548. I checked and my settings are in degrees, not radians.

2) In both my homework (Dynamic Atmosphere) and your examples, when deriving 'y' for example, why do we make '2x' or 'x^2' equal to zero. Why do we not bring them as a constant?
Button navigates to signup pageComment on Brian Conlan's post “2 quick questions: 1) exa...”
(1 vote)
Answer
- Francisco Delgado
  Posted 8 years ago. Direct link to Francisco Delgado's post “If they are stand alone s...”
  If they are stand alone such as x^2+2x+2xy^3. If you are taking the partial derivative with respect to y, you treat the others as a constant. The derivative of a constant is 0, so it becomes
  0+0+2x(3y^2). You'll notice since the last one is multiplied by Y, you treat it as a constant multiplied by the derivative of the function. I.E d/dx 2x^2 is just 2(d/dx(x^2)) or 2(2x) or 4x. Hope this helps.
  Button navigates to signup page
  (2 votes)

Multivariable calculus