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

Step by step, with the reason for every step · Partial derivatives · Calculus 3

Step by step

The question

Question: find the partial derivatives of f(x, y) = x² + y².

f(x, y) = x2 + y2

1 Partial by x

The partial derivative ∂f/∂x is the slope when only x changes.

∂f/∂x: only x changes

For ∂f/∂x, only x changes.

y stays fixed: it acts like a plain number.

It is purple now.

f(x, y) = x2 + y2

Term by term.

x²: the power 2 comes down.

The power goes down by 1.

So 2x.

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

∂f∂x = 2x + 0

Drop the zeros: ∂f/∂x = 2x.

∂f∂x = 2x

Careful!

A fixed letter is not 0!

In x²y, y stays in front.

Only a term with no x at all gives 0.

Careful: fixed y is not 0

2 Partial by y

For ∂f/∂y, only y changes.

x stays fixed: it acts like a plain number.

It is purple now.

f(x, y) = x2 + y2

Term by term.

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

y²: the power 2 comes down.

The power goes down by 1.

So 2y.

∂f∂y = 0 + 2y

Drop the zeros: ∂f/∂y = 2y.

∂f∂y = 2y

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 2.

f(1.001, 2) − f(1, 2)0.001 = 5.002001 − 50.001 ≈ 2.001

And ∂f/∂x at that point is 2.

Almost the same.

Right!

∂f∂x at (1, 2) = 2

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

f(1, 2.001) − f(1, 2)0.001 = 5.004001 − 50.001 ≈ 4.001

And ∂f/∂y at that point is 4.

Almost the same.

Right!

∂f∂y at (1, 2) = 4

Let's look back.

First: partial by x.

Then: partial by y.

Last: check.

Answer
Careful: fixed y is not 0

Your turn

Try this one yourself, then check:

Find the partial derivatives of f(x, y) = 2x2 + 4y2
Show the answer
∂f∂x = 4x, ∂f∂y = 8y

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