For f(x, y) = x^2 + 3xy at (1, 2), what is the directional derivative along (1, 1)/sqrt(2)?

## What is being asked A directional derivative answers: standing at a specific point, if I walk along a specific bearing, how fast does the function change per unit of distance travelled? It reduces a whole surface to one number, given a point and a direction. ## Step 1 — the gradient, symbolically For `f(x, y) = x^2 + 3xy`, differentiate one variable at a time with the other held constant. - With respect to `x`: the term `x^2` contributes `2x`; the term `3xy` is a constant `3y` times `x`, contributing `3y`. So `df/dx = 2x + 3y`. - With respect to `y`: the term `x^2` is constant in `y`, contributing `0`; the term `3xy` is a constant `3x` times `y`, contributing `3x`. So `df/dy = 3x`. Hence `grad f(x, y) = (2x + 3y, 3x)`. ## Step 2 — evaluate at the point At `(x, y) = (1, 2)`: - `df/dx = 2*1 + 3*2 = 2 + 6 = 8` - `df/dy = 3*1 = 3` So `grad f(1, 2) = (8, 3)`. Substituting the point *after* differentiating is essential; substituting first would freeze the function into a constant and destroy the derivative. ## Step 3 — combine with the direction For a differentiable `f` and a direction `u` of length one, the directional derivative is `D_u f = grad f . u = (df/dx) * u1 + (df/dy) * u2` With `u = (1, 1)/sqrt(2)`, i.e. `u1 = u2 = 1/sqrt(2)`: `D_u f(1, 2) = 8 * (1/sqrt(2)) + 3 * (1/sqrt(2)) = 11/sqrt(2)` Rationalised, `11/sqrt(2) = 11*sqrt(2)/2`, numerically about `7.78`. It is positive, so this bearing goes uphill: moving a small distance `s` along it raises `f` by roughly `7.78 * s`. ## Why the direction must have length one The defining limit is `D_u f = lim_{h -> 0} [ f(a + h*u) - f(a) ] / h`. Here `h` is intended to be the distance travelled, which is only true when `u` has length one. Feed in the raw arrow `(1, 1)` instead and you get `8 + 3 = 11`, exactly `sqrt(2)` times the correct value — because `(1, 1)` reaches `sqrt(2)` units of distance for every unit of `h`. Double the arrow to `(2, 2)` and the number doubles again. Without normalisation you are not reporting a property of the bearing at all; you are reporting a property of the particular arrow someone happened to draw. This is the single most common slip on this question, and interviewers watch for whether you normalise unprompted. ## Sanity checks **Upper bound.** No directional derivative can exceed the gradient's magnitude, since `D_u f = |grad f| * cos(theta)` and the cosine is capped at 1. Here `|grad f(1, 2)| = sqrt(64 + 9) = sqrt(73)`, about `8.54`. Our answer `7.78` sits just below it, which is what you would expect: the bearing `(1, 1)/sqrt(2)` is close to, but not exactly, the direction `(8, 3)` in which the steepest rate `sqrt(73)` is achieved. **Sign.** A positive value means uphill along that bearing. Reversing to `(-1, -1)/sqrt(2)` flips the sign to `-11/sqrt(2)`: the same slope walked backwards. **Type.** The directional derivative is a single number, not a vector. If your answer has two components you have confused it with the gradient. **Zero directions.** Any unit direction perpendicular to `(8, 3)` — for instance `(3, -8)/sqrt(73)` — gives `24 - 24 = 0`, no first-order change. ## The shape of the answer Given a point and a direction, the recipe never changes: differentiate symbolically, evaluate at the point, normalise the direction, combine componentwise, then sanity-check the sign and the magnitude bound. Interviewers who ask this are checking arithmetic discipline and whether you understand what the number means, not exotic calculus.