Step by step
The question
Question: find the partial derivatives of f(x, y) = 3x² + 2xy − y³ at (1, 2).
1 Partial by x
The partial derivative ∂f/∂x is the slope when only x changes.
For ∂f/∂x, only x changes.
y stays fixed: it acts like a plain number.
It is purple now.
Term by term.
3x²: the power 2 comes down: 3 times 2 is 6.
The power goes down by 1.
So 6x.
In 2xy, 2y is fixed and waits in front.
x gives 1.
So 2y.
y³ has no x: it is a fixed number, so it gives 0.
Drop the zeros: ∂f/∂x = 6x + 2y.
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 stays fixed: it acts like a plain number.
It is purple now.
Term by term.
3x² has no y: it is a fixed number, so it gives 0.
In 2xy, 2x is fixed and waits in front.
y gives 1.
So 2x.
y³: the power 3 comes down: −1 times 3 is −3.
The power goes down by 1.
So 3y².
Drop the zeros: ∂f/∂y = 2x − 3y².
3 Put the point in
Put x = 1, y = 2 into ∂f/∂x.
Work out each term.
So f_x(1, 2) = 10.
Put x = 1, y = 2 into ∂f/∂y.
Work out each term.
So f_y(1, 2) = −10.
4 Check
Let's check.
Here is the surface z = f(x, y) near (1, 2).
On the orange curve only x changes.
∂f/∂x is its slope.
On the purple curve only y changes.
∂f/∂y is its slope.
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 −10.01.
And ∂f/∂y at that point is −10.
Almost the same.
Right!
Let's look back.
First: partial by x.
Then: partial by y.
Then: put the point in.
Last: check.