Here is Prob. 4, Exercises 8.14, in the book Calculus Vol II by Tom M. Apostol, 2nd edition:
A differentiable scalar field $f$ has, at the point $(1, 2)$, directional derivative $+2$ in the direction toward $(2, 2)$ and $-2$ in the direction toward $(1, 1)$. Determine the gradient vector at $(1, 2)$ and compute the directional derivative in the direction toward $(4, 6)$.
My Attempt:
Let $(a, b)$ be the gradient vector $\nabla f(1, 2)$ of the scalar field $f$ at the point $(1, 2)$.
The vector $\mathbf{u}$ from $(1, 2)$ to $(2, 2)$ is given by $$ \mathbf{u} = (2, 2) - (1, 2) = (1, 0). $$ Now as $f$ is differentiable at $(1, 2)$, so the directional derivative of $f$ in the direction of $\mathbf{u}$ is $$ \nabla f(1, 2) \cdot \mathbf{u} = (a, b) \cdot (1, 0) = a. $$ Therefore we obtain $a = 2$.
The vector $\mathbf{v}$ from $(1, 2)$ to $(1, 1)$ is given by $$ \mathbf{v} = (1, 1) - (1, 2) = (0, -1). $$ Once again as $f$ is differentiable at $(1, 2)$, so the directional derivative of $f$ in the direction of $\mathbf{v}$ is $$ \nabla f(1, 2) \cdot \mathbf{v} = (a, b) \cdot (0, -1) = -b. $$ Therefore we obtain $b = 2$.
Thus the gradient vector of $f$ at $(1, 2)$ is given by $$ \nabla f(1, 2) = (2, 2). $$
Now the vector $\mathbf{w}$ from $(1, 2)$ toward $(4, 6)$ is given by $$ \mathbf{w} = (4, 6)- (1, 2) = (3, 4). $$ So the directional derivative of $f$ at $(1, 2)$ in the direction toward $\mathbf{w}$ is $$ \nabla f(1, 2) \cdot \mathbf{w} = (2, 2) \cdot (3, 4) = 14. $$
Is there any error --- either in logic or calculation --- in this solution?
