Steply

Partial derivatives of f(x, y, z) = xyz + z^2

Step by step, with the reason for every step · Partial derivatives · Calculus 3

Step by step

The question

Question: find the partial derivatives of f(x, y, z) = xyz + z².

f(x, y, z) = xyz + z2

1 Partial by x

The partial derivative ∂f/∂x is the slope when only x changes.

∂f/∂x: only x changes

For ∂f/∂x, only x changes.

y and z stay fixed: they act like plain numbers.

They are purple now.

f(x, y, z) = xyz + z2

Term by term.

In xyz, yz is fixed and waits in front.

x gives 1.

So yz.

z² has no x: it is a fixed number, so it gives 0.

∂f∂x = yz + 0

Drop the zeros: ∂f/∂x = yz.

∂f∂x = yz

Careful!

A fixed letter is not 0!

In x²y, y stays in front.

Only a term with no x at all gives 0.

Careful: fixed y and z is not 0

2 Partial by y

For ∂f/∂y, only y changes.

x and z stay fixed: they act like plain numbers.

They are purple now.

f(x, y, z) = xyz + z2

Term by term.

In xyz, xz is fixed and waits in front.

y gives 1.

So xz.

z² has no y: it is a fixed number, so it gives 0.

∂f∂y = xz + 0

Drop the zeros: ∂f/∂y = xz.

∂f∂y = xz

3 Partial by z

For ∂f/∂z, only z changes.

x and y stay fixed: they act like plain numbers.

They are purple now.

f(x, y, z) = xyz + z2

Term by term.

In xyz, xy is fixed and waits in front.

z gives 1.

So xy.

z²: the power 2 comes down.

The power goes down by 1.

So 2z.

So ∂f/∂z = xy + 2z.

∂f∂z = xy + 2z

4 Check

The slope between two very close points.

Move x by 0.001: the change in f, over 0.001, is about 2.

f(1.001, 2, 1) − f(1, 2, 1)0.001 = 3.002 − 30.001 ≈ 2

And ∂f/∂x at that point is 2.

Almost the same.

Right!

∂f∂x at (1, 2, 1) = 2

Now move y by 0.001: the change in f, over 0.001, is about 1.

f(1, 2.001, 1) − f(1, 2, 1)0.001 = 3.001 − 30.001 ≈ 1

And ∂f/∂y at that point is 1.

Almost the same.

Right!

∂f∂y at (1, 2, 1) = 1

Now move z by 0.001: the change in f, over 0.001, is about 4.

f(1, 2, 1.001) − f(1, 2, 1)0.001 = 3.004001 − 30.001 ≈ 4.001

And ∂f/∂z at that point is 4.

Almost the same.

Right!

∂f∂z at (1, 2, 1) = 4

Let's look back.

First: partial by x.

Then: partial by y.

Then: partial by z.

Last: check.

Answer
Careful: fixed y and z is not 0

Your turn

Try this one yourself, then check:

f(x, y, z) = xyz + 4z2
Show the answer
∂f∂x = yz, ∂f∂y = xz, ∂f∂z = xy + 8z

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