# gradient of f at (1,1) from directional derivatives and vectors

Image of the problem

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

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

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

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

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

• 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). Jan 31 '19 at 20:03
• 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)". Jan 31 '19 at 20:08

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.
• 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 $\nabla f(1,1)= g(1, 1)\vec{i}+ h(1,1)\vec{j}$. Jan 31 '19 at 20:32