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Image of the problem

Find the derivative (gradient) of $f$ at $(1,1)$ given the following directional derivatives where $u=2i+2j$ and $v=3i+j$

$D_uf(1,1)=\frac{3}{\sqrt2}$,

$D_uf(-1,1)=\frac{7}{\sqrt2}$,

$D_vf(1,1)=-\frac{1}{\sqrt10}$,

$D_vf(-1,1)=-\frac{5}{\sqrt10}$.

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  • $\begingroup$ For the task of finding the gradient at (1, 1), information about the function and its derivatives at (-1, -1) is irrelevant. The function in a neighborhood of (-1, -1) might be completely different from the function in a neighborhood of (1, 1). $\endgroup$
    – user247327
    Jan 31 '19 at 20:03
  • $\begingroup$ Also there must be more information about f. f can be continuous at (1,1) and have those directional derivatives and NOT have a gradient. You need that the function is "differentiable in a neighborhood of (1, 1)". $\endgroup$
    – user247327
    Jan 31 '19 at 20:08
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Assuming that this function has a gradient (see my comments) we can write it as $\nabla f= g(x,y)\vec{i}+ h(x,y)\vec{j}$. Now, its derivative in the direction of $\vec{u}= \vec{i}+ 2\vec{j}$ is $2g(1,1)+ 2h(1,1)= \frac{3}{\sqrt{2}}$ and its derivative in the direction of $\vec{v}= 3\vec{i}+ \vec{j}$ is $3g(1,1)+ h(1,1)= \frac{7}{\sqrt{2}}$. That gives two equations that can be solved for g(1, 1) and h(1, 1). But as I implied in my comments, the gradient is local. Knowing the value of g and h at (1, 1) does not tell us what g and h (and so f) are for other points.

Were you given other information about f, like it being linear of a conic?

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  • $\begingroup$ I have included a picture of the problem I've been given $\endgroup$
    – Zaltah
    Jan 31 '19 at 20:26
  • $\begingroup$ Ah, so the problem is just asking for the gradient at (1, 1). For some reason I missed that when I first read the problem. All you need to do is solve $2g(1, 1)+ 2h(1,1)= \frac{3}{\sqrt{2}}$ and $3g(1.1)+ h(1,1)= \frac{7}{\sqrt{2}}$ for g(1,1) and h(1,1). Then $\nabla f(1,1)= g(1, 1)\vec{i}+ h(1,1)\vec{j}$. $\endgroup$
    – user247327
    Jan 31 '19 at 20:32

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