Step by step
The question
Question: find the partial derivatives of f(x, y) = 2xy³ − x² at (3, −1).
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.
In 2xy³, 2y³ is fixed and waits in front.
x gives 1.
So 2y³.
x²: the power 2 comes down: −1 times 2 is −2.
The power goes down by 1.
So 2x.
So ∂f/∂x = 2y³ − 2x.
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.
In 2xy³, 2x is fixed and waits in front.
y³ gives 3y².
So 6xy².
x² has no y: it is a fixed number, so it gives 0.
Drop the zeros: ∂f/∂y = 6xy².
3 Put the point in
Put x = 3, y = −1 into ∂f/∂x.
Work out each term.
So f_x(3, −1) = −8.
Put x = 3, y = −1 into ∂f/∂y.
So f_y(3, −1) = 18.
4 Check
Let's check.
Here is the surface z = f(x, y) near (3, −1).
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 −8.
And ∂f/∂x at that point is −8.
Almost the same.
Right!
Now move y by 0.001: the change in f, over 0.001, is about 17.98.
And ∂f/∂y at that point is 18.
Almost the same.
Right!
Let's look back.
First: partial by x.
Then: partial by y.
Then: put the point in.
Last: check.