Second partial derivatives of f(x, y) = x^2 y + 3xy^2

Step by step

The question

Question: find the second partial derivatives of f(x, y) = x²y + 3xy².

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 x²y, y is fixed and waits in front.

x² gives 2x.

So 2xy.

In 3xy², 3y² is fixed and waits in front.

x gives 1.

So 3y².

So ∂f/∂x = 2xy + 3y².

2 Partial by y

Term by term.

In x²y, x² is fixed and waits in front.

y gives 1.

So x².

In 3xy², 3x is fixed and waits in front.

y² gives 2y.

So 6xy.

So ∂f/∂y = x² + 6xy.

3 The second partials

Now take each answer's partial derivatives again.

f_xy means: first by x, then by y.

Term by term.

In 2xy, 2y is fixed and waits in front.

x gives 1.

So 2y.

3y² has no x: it is a fixed number, so it gives 0.

Drop the zeros: f_xx = 2y.

Term by term.

In 2xy, 2x is fixed and waits in front.

y gives 1.

So 2x.

3y²: the power 2 comes down: 3 times 2 is 6.

The power goes down by 1.

So 6y.

So f_xy = 2x + 6y.

Term by term.

x² has no y: it is a fixed number, so it gives 0.

In 6xy, 6x is fixed and waits in front.

y gives 1.

So 6x.

Drop the zeros: f_yy = 6x.

Term by term.

x²: the power 2 comes down.

The power goes down by 1.

So 2x.

In 6xy, 6y is fixed and waits in front.

x gives 1.

So 6y.

So f_yx = 2x + 6y.

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.