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Let the directional derivative of a function $f(x,y)$ at a point $P$ in the direction of $\big(\frac{1}{\sqrt{5}}\,i+\frac{2}{\sqrt{5}}\,j\big)$ be $\frac{16}{\sqrt{5}}$ and $\frac{\partial f}{\partial x}$ evaluated at $P$ be 6. What is the directional derivative in the direction of $i - j$?

Looking up numerous examples, I still can't find out what to do with all the information given.

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I'm assuming that you are asking for the directional derivative in the new direction evaluated at the same point. So we are given:

$$\nabla f \cdot \hat{u} \Big|_P = \frac{16}{\sqrt{5}}$$

$$\hat{u} = \begin{pmatrix} \frac{1}{\sqrt{5}} \\ \frac{2}{\sqrt{5}} \end{pmatrix}$$

$$f_x \Big|_P = 6$$

So using the first piece of information:

$$f_x \Big|_P u_1 + f_y \Big|_P u_2 = \frac{16}{\sqrt{5}}$$

$$(6)\Big( \frac{1}{\sqrt{5}} \Big) + f_y \Big|_P \Big( \frac{2}{\sqrt{5}} \Big) = \frac{16}{\sqrt{5}}$$

$$f_y \Big|_P = 5$$

With this we define the new direction:

$$v = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$$

$$\hat{v} = \begin{pmatrix} \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} \end{pmatrix}$$

and compute the directional derivative:

$$\nabla f \cdot \hat{v} \Big|_P = f_x \Big|_P v_1 + f_y \Big|_P v_2$$

$$= (6) \Big( \frac{1}{\sqrt{2}} \Big) + (5) \Big( -\frac{1}{\sqrt{2}} \Big)$$

$$= \frac{1}{\sqrt{2}}$$

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Hint:

$\nabla{f}\cdot (\frac{1}{\sqrt5}, \frac{2}{\sqrt5}) = (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}) = (6)(\frac{1}{\sqrt5}) + (\frac{\partial f}{\partial y})(\frac{2}{\sqrt5}) = \frac{16}{\sqrt5}$

Thus, you can solve for $\frac{\partial f}{\partial y}$

With this, you just need to compute $\nabla{f}\cdot (\frac{1}{\sqrt2},\frac{-1}{\sqrt2})$ at $P$, which is certainly possible, because you now know $\frac{\partial f}{\partial y}$ at $P$

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