Step by step
The question
Question: find the second partial derivatives of f(x, y) = 4x³ − 2xy + y⁴.
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.
4x³: the power 3 comes down: 4 times 3 is 12.
The power goes down by 1.
So 12x².
In 2xy, 2y is fixed and waits in front.
x gives 1.
So 2y.
y⁴ has no x: it is a fixed number, so it gives 0.
Drop the zeros: ∂f/∂x = 12x² − 2y.
2 Partial by y
Term by term.
4x³ 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.
y⁴: the power 4 comes down.
The power goes down by 1.
So 4y³.
Drop the zeros: ∂f/∂y = −2x + 4y³.
3 The second partials
Now take each answer's partial derivatives again.
f_xy means: first by x, then by y.
Term by term.
12x²: the power 2 comes down: 12 times 2 is 24.
The power goes down by 1.
So 24x.
2y has no x: it is a fixed number, so it gives 0.
Drop the zeros: f_xx = 24x.
Term by term.
12x² has no y: it is a fixed number, so it gives 0.
2y gives 2.
Drop the zeros: f_xy = −2.
Term by term.
2x has no y: it is a fixed number, so it gives 0.
4y³: the power 3 comes down: 4 times 3 is 12.
The power goes down by 1.
So 12y².
Drop the zeros: f_yy = 12y².
Term by term.
2x gives 2.
4y³ has no x: it is a fixed number, so it gives 0.
Drop the zeros: f_yx = −2.
4 Check
Let's check.
The two mixed ones, f_xy and f_yx, must come out the same.
They do.
Right!
Let's look back.
First: partial by x.
Then: partial by y.
Then: the second partials.
Last: check.