Directional derivative of f(x, y) = x^2 y at (1, 3) in the direction… Step by Step

Step by step

The question

Question: find the directional derivative of f(x, y) = x²y at (1, 3), in the direction of ⟨0, 1⟩.

f(x, y) = x²y

1 The partial derivatives

Term by term.

In x²y, y is fixed and waits in front.

x² gives 2x.

So 2xy.

So ∂f/∂x = 2xy.

∂f∂x = 2xy

Term by term.

In x²y, x² is fixed and waits in front.

y gives 1.

So x².

So ∂f/∂y = x².

∂f∂y = x²

2 Find the gradient

The gradient ∇f puts the partial derivatives together as one vector.

∇f = ⟨∂f∂x, ∂f∂y⟩

So ∇f = ⟨2xy, x²⟩.

∇f = ⟨2xy, x²⟩

At the point, put x = 1 and y = 3 in: ∇f(1, 3) = ⟨6, 1⟩.

∇f(1, 3) = ⟨6, 1⟩

3 Make it a unit vector

⟨0, 1⟩ already has length 1.

u = ⟨0, 1⟩

4 Dot with the gradient

The slope in direction u is the gradient dot u.

D_uf = ∇f • u

Multiply the matching parts and add.

D_uf = 6 × 0 + 1 × 1

Careful!

With ⟨0, 1⟩ itself, the answer would be 1 times too big.

Careful: use the unit vector u

6 times 0 is 0.

1 times 1 is 1.

D_uf = 0 + 1

Add: the slope is 1.

D_uf = 1

5 Check

Let's check: take a tiny step of 0.001 along u.

The change in f/0.001 is about 1.

f(1, 3.001) − f(1, 3)0.001 ≈ 1

Our answer 1 is 1.

Almost the same.

Right!

1 = 1

Let's look back.

First: the partial derivatives.

Then: find the gradient.

Then: make it a unit vector.

Then: dot with the gradient.

Last: check.

Answer