Discussion
The directional derivative of \( \phi \) in the direction of a unit vector \( \mathbf{u} \) is
Answer: A
The rate of change along a direction is the component of the gradient in that direction, so
\[ \frac{d\phi}{ds} \;=\; \nabla\phi \cdot \mathbf{u} \;=\; |\nabla\phi|\cos\theta. \]
A cross product would give a vector rather than a rate, and the product of magnitudes discards the angle, which is exactly what decides how much of the gradient the direction picks up
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