Partial Derivatives

Derivatives where we treat other variables as constants.


Here is a function of one variable (x):

f(x) = x2

And its derivative (using the Power Rule) is:

f’(x) = 2x


But what about a function of two variables (x and y):

f(x,y) = x2 + y3

To find its partial derivative with respect to x we treat y as a constant (imagine y is a number like 7 or something):

f’x = 2x + 0 = 2x


To find the partial derivative with respect to y, we treat x as a constant:

f’y = 0 + 3y2 = 3y2



That is all there is to it. Just remember to treat all other variables as if they are constants.


Holding A Variable Constant

So what does "holding a variable constant" look like?

Cylinder Dimensions

Example: the volume of a cylinder is V = π r2 h

We can write that in "multi variable" form as

f(r,h) = π r2 h


For the partial derivative with respect to r we hold h constant, and r changes:

Cylinder with r changing

f’r = π (2r) h = 2πrh

(The derivative of r2 with respect to r is 2r, and π and h are constants)

It says "as only the radius changes (by the tiniest amount), the volume changes by 2πrh"

It is like we add a skin with a circle's circumference (2πr) and a height of h.


For the partial derivative with respect to h we hold r constant:

Cylinder with r changing

f’h = π r2 (1)= πr2

(π and r2 are constants, and the derivative of h with respect to h is 1)

It says "as only the height changes (by the tiniest amount), the volume changes by πr2"

It is like we add the thinnest disk on top with a circle's area of πr2.

Let's see another example.

Example: The surface area of a square prism.

Cylinder with r changing

The surface is: the top and bottom with areas of x2 each, and 4 sides of area xy:

f(x,y) = 2x2 + 4xy

fx = 4x + 4y

fy = 0 + 4x = 4x

Three or More Variables

We can have 3 or more variables. Just find the partial derivative of each variable in turn while treating all other variables as constants.

Example: The volume of a cube with a square prism cut out from it.

Cylinder with r changing

f(x,y,z) = z3 − x2y

f’x = 0 − 2xy = −2xy

f’y = 0 − x2 = −x2

f’z = 3z2 − 0 = 3z2

When there are many x's and y's it can get confusing, so a mental trick is to change the "constant" variables into letters like "c" or "k" that look like constants.

Example: f(x,y) = y3sin(x) + x2tan(y)

It has x's and y's all over the place! So let us try the letter change trick.

With respect to x we can change "y" to "k":

f(x,y) = k3sin(x) + x2tan(k)

f’x = k3cos(x) + 2x tan(k)

But remember to turn it back again!

f’x = y3cos(x) + 2x tan(y)

Likewise with respect to y we turn the "x" into a "k":

f(x,y) = y3sin(k) + k2tan(y)

f’y = 3y2sin(k) + k2sec2(y)

f’y = 3y2sin(x) + x2sec2(y)

But only do this if you have trouble remembering, as it is a little extra work.


Notation: here we use f’x to mean "the partial derivative with respect to x", but another very common notation is to use a funny backwards d (∂) like this:

∂f∂x = 2x

Which is the same as:

f’x = 2x

∂ is called "del" or "dee" or "curly dee"

So ∂f ∂x is said "del f del x"

Example: find the partial derivatives of f(x,y,z) = x4 − 3xyz using "curly dee" notation

f(x,y,z) = x4 − 3xyz

∂f∂x = 4x3 − 3yz

∂f∂y = −3xz

∂f∂z = −3xy

You might prefer that notation, it certainly looks cool.