Step by step
The question
Question: find the partial derivatives of f(x, y, z) = 2x²z − 3xy + z at (1, 2, −1).
1 Partial by x
The partial derivative ∂f/∂x is the slope when only x changes.
For ∂f/∂x, only x changes.
y and z stay fixed: they act like plain numbers.
They are purple now.
Term by term.
In 2x²z, 2z is fixed and waits in front.
x² gives 2x.
So 4xz.
In 3xy, 3y is fixed and waits in front.
x gives 1.
So 3y.
z has no x: it is a fixed number, so it gives 0.
Drop the zeros: ∂f/∂x = 4xz − 3y.
Careful!
A fixed letter is not 0!
In x²y, y stays in front.
Only a term with no x at all gives 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.
Term by term.
2x²z has no y: it is a fixed number, so it gives 0.
In 3xy, 3x is fixed and waits in front.
y gives 1.
So 3x.
z has no y: it is a fixed number, so it gives 0.
Drop the zeros: ∂f/∂y = −3x.
3 Partial by z
For ∂f/∂z, only z changes.
x and y stay fixed: they act like plain numbers.
They are purple now.
Term by term.
In 2x²z, 2x² is fixed and waits in front.
z gives 1.
So 2x².
3xy has no z: it is a fixed number, so it gives 0.
z gives 1.
Drop the zeros: ∂f/∂z = 2x² + 1.
4 Put the point in
Put x = 1, y = 2, z = −1 into ∂f/∂x.
Work out each term.
So f_x(1, 2, −1) = −10.
Put x = 1, y = 2, z = −1 into ∂f/∂y.
So f_y(1, 2, −1) = −3.
Put x = 1, y = 2, z = −1 into ∂f/∂z.
Work out each term.
So f_z(1, 2, −1) = 3.
5 Check
The slope between two very close points.
Move x by 0.001: the change in f, over 0.001, is about −10.
And ∂f/∂x at that point is −10.
Almost the same.
Right!
Now move y by 0.001: the change in f, over 0.001, is about −3.
And ∂f/∂y at that point is −3.
Almost the same.
Right!
Now move z by 0.001: the change in f, over 0.001, is about 3.
And ∂f/∂z at that point is 3.
Almost the same.
Right!
Let's look back.
First: partial by x.
Then: partial by y.
Then: partial by z.
Then: put the point in.
Last: check.