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

Step by step

The question

Question: find the directional derivative of f(x, y) = xy² at (2, 1), in the direction of ⟨3, -4⟩.

1 The partial derivatives

Term by term.

In xy², y² is fixed and waits in front.

x gives 1.

So y².

So ∂f/∂x = y².

Term by term.

In xy², x is fixed and waits in front.

y² gives 2y.

So 2xy.

So ∂f/∂y = 2xy.

2 Find the gradient

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

So ∇f = ⟨y², 2xy⟩.

At the point, put x = 2 and y = 1 in: ∇f(2, 1) = ⟨1, 4⟩.

3 Make it a unit vector

The direction ⟨3, -4⟩ has length 5.

A direction must have length 1.

So divide each part by 5: u = ⟨3/5, −4/5⟩.

4 Dot with the gradient

The slope in direction u is the gradient dot u.

Multiply the matching parts and add.

Careful!

With ⟨3, -4⟩ itself, the answer would be 5 times too big.

1 times 3/5 is 3/5.

4 times −4/5 is −16/5.

Add: the slope is −13/5.

5 Check

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

The change in f/0.001 is about −2.6.

Our answer −13/5 is −2.6.

Almost the same.

Right!

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.