Gradient of sin(x + 2y + 3z): A Fixed Direction

One outer derivative, three inner derivatives

Set $u = x + 2y + 3z$. The outer function is $\sin u$, whose derivative is $\cos u$ no matter which variable you are differentiating. What changes from component to component is the inner derivative.

Multiply the shared outer factor by each inner derivative in turn.

Factoring the cosine out shows the structure at a glance. The gradient is always a scalar multiple of $\left\langle 1, 2, 3 \right\rangle$, so its direction is fixed up to sign for the whole of space.

The mistake: losing the coefficients 2 and 3

Writing $\nabla f = \left\langle \cos u, \cos u, \cos u \right\rangle$ is the standard error. It comes from treating $\frac{\partial}{\partial y}\sin(u)$ as $\cos(u)$ and stopping, which skips the inner derivative. Those coefficients are the whole point: $f$ responds three times as strongly to a change in $z$ as to the same change in $x$.

• Rewrite the argument first if it helps: $\sin(x + 2y + 3z)$ has inner function $u = x+2y+3z$, and $\frac{\partial u}{\partial y} = 2$ is not 1.
• A quick test: at the origin the true rate in the $z$ direction is 3, so a small step of 0.01 in $z$ changes $f$ by about 0.03, not 0.01.
• The reverse error, attaching the coefficients to the argument as in $\cos(2y)$, changes the function rather than the derivative.

Level surfaces are parallel planes

At the origin, $u = 0$ and $\cos 0 = 1$, so the gradient takes its largest possible size.

The level surfaces $\sin(x+2y+3z) = c$ are stacks of parallel planes $x + 2y + 3z = \text{constant}$, and $\left\langle 1,2,3 \right\rangle$ is the normal vector to those planes. This matches the general rule that the gradient is perpendicular to level surfaces, and here it is visible directly in the coefficients.

The gradient vanishes where $\cos(x+2y+3z) = 0$, that is on the planes $x + 2y + 3z = \frac{\pi}{2} + k\pi$. Those are the crests and troughs of the wave, entire planes of critical points where $f = \pm 1$.