Maximum Directional Derivative

Summary

Among all unit directions, the directional derivative is maximized in the direction of the gradient. That maximum value equals the gradient’s magnitude $\parallel \mathrm{\nabla }f\parallel$ . The steepest descent direction is $-\mathrm{\nabla }f/\parallel \mathrm{\nabla }f\parallel$ .

Prerequisites

Directional Derivative

Theorem

If $f$ is differentiable at $\mathbf{a}$ and $\mathrm{\nabla }f(\mathbf{a})\ne \mathbf{0}$ , then for unit vectors $\mathbf{u}$ ,

where $\theta$ is the angle between $\mathrm{\nabla }f(\mathbf{a})$ and $\mathbf{u}$ . Therefore

attained at $\mathbf{u}=\mathrm{\nabla }f(\mathbf{a})/\parallel \mathrm{\nabla }f(\mathbf{a})\parallel$ , and

Notation Caution

Do not write $|D_{\mathbf{u}}f(\mathbf{a})|$ for the maximum rate. The absolute value $|D_{\mathbf{u}}f|$ is the magnitude of the rate in a fixed direction $\mathbf{u}$ ; the maximum over directions is $\parallel \mathrm{\nabla }f\parallel$ .

Conditions / Assumptions

• Differentiability at the point (so $D_{\mathbf{u}}f=\mathrm{\nabla }f\cdot \mathbf{u}$ ).
• If $\mathrm{\nabla }f(\mathbf{a})=\mathbf{0}$ , every directional derivative is zero (critical point).

Worked Example

For $f(x,y)=x^2+y^2$ at $(1,1)$ , $\mathrm{\nabla }f(1,1)=(2,2)$ and

The maximum directional derivative is $2\sqrt{2}$ in the direction $(1/\sqrt{2},1/\sqrt{2})$ .

Common Mistakes

• Using $|D_{\mathbf{u}}f|$ as if it meant “max over $\mathbf{u}$ .”
• Forgetting to normalize the gradient when stating the maximizing direction.

Connections

• Directional Derivative, Maxima and Minima, optimization with constraints Lagrange Multipliers

References

The steepest-ascent theorem is standard in multivariable calculus.^[1]