Partial derivatives of f(x, y) = x^2 y + 3y Easy

The question

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

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.

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

Drop the zeros: ∂f/∂x = 2xy.

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 x²y, x² is fixed and waits in front.

y gives 1.

So x².

3y gives 3.

So ∂f/∂y = x² + 3.

3 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 4.

And ∂f/∂x at that point is 4.

Almost the same.

Right!

Now move y by 0.001: the change in f, over 0.001, is about 4.

And ∂f/∂y at that point is 4.

Almost the same.

Right!

Let's look back.

First: partial by x.

Then: partial by y.

Last: check.

▶ Watch it solved

Partial derivatives of f(x, y) = 3x^2 + 2xy - y^3 at (1, 2) Medium

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.

▶ Watch it solved

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

The question

Question: find the second partial derivatives of f(x, y) = x³y².

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 x³y², y² is fixed and waits in front.

x³ gives 3x².

So 3x²y².

So ∂f/∂x = 3x²y².

2 Partial by y

Term by term.

In x³y², x³ is fixed and waits in front.

y² gives 2y.

So 2x³y.

So ∂f/∂y = 2x³y.

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 3x²y², 3y² is fixed and waits in front.

x² gives 2x.

So 6xy².

So f_xx = 6xy².

Term by term.

In 3x²y², 3x² is fixed and waits in front.

y² gives 2y.

So 6x²y.

So f_xy = 6x²y.

Term by term.

In 2x³y, 2x³ is fixed and waits in front.

y gives 1.

So 2x³.

So f_yy = 2x³.

Term by term.

In 2x³y, 2y is fixed and waits in front.

x³ gives 3x².

So 6x²y.

So f_yx = 6x²y.

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.

▶ Watch it solved