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I've been asked to find the derivative of $f(\vec x)$ at a point $\vec a$ in the direction of $\vec v=$$\begin{bmatrix}-2\\3\end{bmatrix}$$ $. The information I'm given to figure this out is as follows:

$f(\vec a)$ in the direction of $\vec v_1=$$\begin{bmatrix}1\\1\end{bmatrix}$$ $ is $\frac{\sqrt 2} 2 + \pi$

$f(\vec a)$ in the direction of $\vec v_2=$$\begin{bmatrix}1\\2\end{bmatrix}$$ $ is $\frac{1+2\pi\sqrt 2} {\sqrt 5}$

The only hint they give me is to write $\nabla f(\vec a)=$$\begin{bmatrix}b\\c\end{bmatrix}$$ $ and then use the given solved directional derivatives to find the constants $b$ and $c$.

I know how to find a directional derivative at a point, but this question seems to be going backwards and I can't get the math to work out. I gathered from another similar question here that this can be solved with a linear combination, but it's not explained how to do that. We've been learning about inner products as well along with directional derivatives, which seems to fit into this problem but doesn't help me much.

How can I find a directional derivative given derivatives in different directions, possibly using a linear combination? If a linear combination is not needed, how can I calculate the directional derivative in the direction $\vec v=$$\begin{bmatrix}-2\\3\end{bmatrix}$$ $?

Thanks in advance!

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    $\begingroup$ You can write $v$ as a linear combination of $v_1$ and $v_2$, then use linearity of matrices. $\endgroup$ – Dave Sep 27 '18 at 1:47
  • $\begingroup$ You need the additional assumption that $f$ is differentiable at $a$, else there’s no way to solve the problem. $\endgroup$ – amd Sep 27 '18 at 3:10
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For real valued functions, the derivative can be interpreted as the slope, which is just a number, so we interpret the derivative at a point as a number. We could picture the derivative at a point as the tangent line to that point. For functions of several variables, the derivative is not a single number. Rather, we define the derivative to be a linear function, i.e., the tangent plane to a point, or the tangent $n$-space to a point.

When the derivative exists for a function $\Bbb R^2\to \Bbb R$, it is a tangent plane. Thus knowing the derivative along two linearly independent vectors tells us the derivative in any direction. If $D_a(v)$ is the derivative of $f$ at $a$ in the direction $v$, then since $D$ is linear,

$$D_a\left(\begin{bmatrix}-2 \\ 3 \end{bmatrix} \right)=D_a\left(-7\begin{bmatrix}1 \\ 1 \end{bmatrix}+5\begin{bmatrix}1 \\ 2 \end{bmatrix} \right)=-7D_a\left(\begin{bmatrix}1 \\ 1 \end{bmatrix}\right)+5D_a\left(\begin{bmatrix}1 \\ 2 \end{bmatrix}\right).$$

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So $v = 5v_2 - 7v_1$, so surely the directional derivative at $a$ in the direction of $v$ is just $5(\frac{1+2\pi\sqrt 2} {\sqrt 5})-7(\frac{\sqrt 2} 2 + \pi)$

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