# Gradient with v_1:=1/(\sqrt(2))((1),(1)) and v_2:=1/(5)((4),(-3)) given

Let $g:{\mathbb{R}}^{\mathbb{2}}\to \mathbb{R}$ be totally differentiable in the point $a\in {\mathbb{R}}^{\mathbb{2}}$
Let ${v}_{1}:=\frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\ 1\end{array}\right)$ and ${v}_{2}:=\frac{1}{5}\left(\begin{array}{c}4\\ -3\end{array}\right)$ be normalized direction vectors in ${\mathbb{R}}^{\mathbb{2}}$
It is $\frac{\mathrm{\partial }g}{\mathrm{\partial }{v}_{1}}\left(a\right)=5\sqrt{2}$ and $\frac{\mathrm{\partial }g}{\mathrm{\partial }{v}_{2}}\left(a\right)=1$
With the info given above, how can one find out $\mathrm{\nabla }g\left(a\right)$?
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Gauge Howard
Given that ${v}_{1}$ and ${v}_{2}$ are unit vectors, the the direction derivative gives
${\mathrm{\nabla }}_{{v}_{2}}g=\mathrm{\nabla }g\cdot {v}_{2}$
and
${\mathrm{\nabla }}_{{v}_{2}}g=\mathrm{\nabla }g\cdot {v}_{2}$
Use these to build a linear system of 2 equations in the 2 components of $\mathrm{\nabla }g$, and solve. So, if
$\mathrm{\nabla }g=\left[x,y{\right]}^{T}$, then
$\frac{1}{\sqrt{2}}x+\frac{1}{\sqrt{2}}y=5\sqrt{2}$
and
$\frac{4}{5}x-\frac{3}{5}y=1$
After multiplying the first equation through by $\sqrt{2}$ and second equation by 5, they becomes
$x+y=10$
$4x-3y=5$
It is trivial to solve to get x=5 , y=5
Therefore, $\mathrm{\nabla }g=\left[\begin{array}{c}5\\ 5\end{array}\right]$