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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?

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I think you were lucky for the first part because you got unitary vectors. The directional derivative is defined for directions, so is, we have to use unitary vectors. So, the last part has to be:

$$\nabla f(1, 2) \cdot \dfrac{\mathbf{w}}{\left|\mathbf{w}\right|} = (2, 2) \cdot \dfrac{1}{5}(3, 4) = \dfrac{14}{5}$$

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