Multivariable calculus

Directional derivative of x e^y at (3, 0) along (0, 1)

At (3, 0) the gradient of f(x,y) = x e^y is (1, 3). The direction (0, 1) is already a unit vector pointing along the y axis, so the dot product picks out the y partial alone and the directional derivative is 3. Axis directions always recover ordinary partial derivatives.

f(x,y)=x eyf(x,y) = x\,e^{y}
Duf(3,0)=∂f∂y(3,0)=3D_{\mathbf{u}} f(3,0) = \frac{\partial f}{\partial y}(3,0) = 3

At x = 3, y = 0, in the unit direction (0, 1).

A partial derivative is a directional derivative

Holding yy fixed makes eye^{y} a constant, so ∂f∂x=ey\frac{\partial f}{\partial x} = e^{y}. Holding xx fixed leaves xx as the constant multiplier on eye^{y}, so ∂f∂y=x ey\frac{\partial f}{\partial y} = x\,e^{y}, which is the original function again.

∇f(3,0)=⟨e0,  3e0⟩=⟨1,3⟩\nabla f(3,0) = \left\langle e^{0},\; 3e^{0} \right\rangle = \langle 1, 3 \rangle

Now dot with u=⟨0,1⟩\mathbf{u} = \langle 0, 1 \rangle. The zero kills the xx term and the one keeps the yy term untouched, which is exactly the definition of the partial with respect to yy.

D⟨0,1⟩f=0⋅∂f∂x+1⋅∂f∂y=∂f∂y(3,0)=3D_{\langle 0, 1 \rangle} f = 0\cdot\frac{\partial f}{\partial x} + 1\cdot\frac{\partial f}{\partial y} = \frac{\partial f}{\partial y}(3,0) = 3

The mistake: treating a partial as a different object

Directional derivatives are usually taught after partials, so they can look like a separate machine. They are the same measurement: the partials are the two directional derivatives along ⟨1,0⟩\langle 1, 0 \rangle and ⟨0,1⟩\langle 0, 1 \rangle, and every other direction is a weighted blend of them.

  • Direction ⟨1,0⟩\langle 1, 0 \rangle gives ∂f∂x(3,0)=1\frac{\partial f}{\partial x}(3,0) = 1
  • Direction ⟨0,1⟩\langle 0, 1 \rangle gives ∂f∂y(3,0)=3\frac{\partial f}{\partial y}(3,0) = 3
  • Direction ⟨0.6,0.8⟩\langle 0.6, 0.8 \rangle gives 1(0.6)+3(0.8)=3.01(0.6) + 3(0.8) = 3.0, which coincidentally ties the yy axis rate at this point

The related slip is direction ⟨0,2⟩\langle 0, 2 \rangle, which points the same way but has length 2 and would report 6. And dropping the xx when differentiating x eyx\,e^{y} with respect to yy gives 1 instead of 3.

Not the steepest direction, but close

The steepest ascent at (3,0)(3,0) runs along ⟨1,3⟩/10≈⟨0.316,0.949⟩\langle 1, 3 \rangle / \sqrt{10} \approx \langle 0.316, 0.949 \rangle, with rate 10≈3.162\sqrt{10} \approx 3.162.

cos⁡θ=310≈0.9487,θ≈18.4∘\cos\theta = \frac{3}{\sqrt{10}} \approx 0.9487, \qquad \theta \approx 18.4^{\circ}

So the yy axis captures about 95 percent of the maximum available rate at this point. That is the usual pattern when one partial is much larger than the other: its axis is nearly the steepest direction.

The build re-checks that (0, 1) has length 1, rebuilds the gradient of x e^y at (3, 0) by central differences on the function itself, and fails unless the dot product is still 3.

Frequently asked questions

Is the answer 3 just the partial derivative with respect to y?

Yes. Along the unit vector ⟨0,1⟩\langle 0, 1 \rangle the dot product reduces to ∂f∂y(3,0)=3e0=3\frac{\partial f}{\partial y}(3,0) = 3e^{0} = 3. Partials are the axis cases of directional derivatives.

Why is the y partial larger than the x partial at (3, 0)?

Because ∂f∂y=x ey\frac{\partial f}{\partial y} = x\,e^{y} scales with xx, and x=3x = 3 there, while ∂f∂x=ey\frac{\partial f}{\partial x} = e^{y} does not depend on xx at all and equals 1.