Finding directional derivative given directional derivative in another direction I know how to find the directional derivative given a function but how do I find the derivative given a directional derivative is another direction?

 A: Given a function $f(x)$ and its gradient $g=\nabla f,\,$ we can calculate the derivatives of $f$ in the direction of two distinct unit vectors $(a,b)$
$$\eqalign{
\alpha &= g^Ta,\quad\beta &= g^Tb,\qquad \{a,b,g\}\in{\mathbb R}^{2}
}$$
These vectors form a basis for ${\mathbb R}^{2}$ so any other unit vector can be written as a linear combination.
$$\eqalign{
c &= \lambda a + \mu b \\
}$$
Multiplying by our two basis vectors yields a linear system which can be solved for $(\lambda,\mu)$
$$
\begin{array}{rr}
  \lambda&+&(a^Tb)\,\mu &= &a^Tc\\
  (b^Ta)\,\lambda&+&\mu &= &b^Tc \\
\end{array}
\\
$$
while multiplying by the gradient, yields the answer to your question:
$$\eqalign{
\gamma &= g^Tc \;=\; \lambda\alpha + \mu\beta \\
}$$
A: There is a theorem that says the following:
Suppose that $f(x,y)$ is differentiable and if $v=(v_1,v_2)$ is a unit vector then
$ D_v f(x,y) = v_1 \frac{\partial f}{\partial x}(x,y)  + v_2\frac{\partial f}{\partial y}(x,y) $
Here $D_vf(x,y)$ means the directional derivative of $f$ in direction $v$ and the point $(x,y)$. If you accept this then you can calculate the directional derivative you want by writing it as a linear combination of the directional derivatives you are given.
As to a proof of the above theorem, write down the definition of directional derivative:
$D_v f(x,y) = \lim_{h\rightarrow 0} \frac{f(x+hv_1,y+hv_2)-f(x,y)}{h} = \lim_{h\rightarrow 0} \frac{f(x+hv_1,y+hv_2) - f(x+hv_1,y) + f(x+hv_1,y)-f(x,y)}{h}$
$ = v_2\lim_{h\rightarrow 0} \frac{f(x+hv_1,y+hv_2)-f(x+hv_1,y)}{v_2h} + v_1\lim_{h\rightarrow 0} \frac{f(x+hv_1,y)-f(x,y)}{v_1h} = v_1 \frac{\partial f}{\partial x}(x,y)  + v_2\frac{\partial f}{\partial y}(x,y)$
A: Hint. Just proceed. If the gradient of $f$ be $(f_x,f_y),$ then we have that the product $$(f_x,f_y)\cdot \frac{1}{\sqrt 2}(1,1)=2\sqrt 2.$$ Similarly we have that $$(f_x,f_y)\cdot \frac{1}{2}(0,-2)=-3,$$ where $f_x,f_y$ are the partials of $f$ evaluated at $(1,2).$ This gives a linear system which may be solved for $f_x,f_y.$ Then the derivative sought is given by $$(f_x,f_y)\cdot \frac{1}{\sqrt {25}}(-1,-2).$$
