Q:

Prove that: $\nabla f(x,y)=\frac{\partial f(x,y)}{\partial \vec{u}}\vec{u}+ \frac{\partial f(x,y)}{\partial \vec{v}}\vec{v}$

Let $ f $ be differentiable and be $\vec{u}$ and $\vec{v}$ two vectors of $\mathbb {R}^2$, unit and orthogonal. Prove that: $$\nabla f(x,y)=\frac{\partial f(x,y)}{\partial \vec{u}}\vec{u}+ \frac{\partial f(x,y)}{\partial \vec{v}}\vec{v}$$
$\mathcal{Proof}$: Let's make $\vec{u} = (u_1, u_2) $ and $\vec{v} = (v_1, v_2)$. Then by hypothesis it is fulfilled that:
$i)$ $u_1v_1+u_2v_2=0$
$ii)$ $u_1^2+u_2^2=1=v_1^2+v_2^2$
Also as $ f $ is differentiable:
$iii)$ $\frac{\partial f(x,y)}{\partial \vec{u}}=\nabla f(x,y)\cdot \vec{u}$ 
$iv)$ $\frac{\partial f(x,y)}{\partial \vec{v}}=\nabla f(x,y)\cdot \vec{v}$

Then $\frac{\partial f(x,y)}{\partial \vec{u}}\vec{u}+ \frac{\partial f(x,y)}{\partial \vec{v}}\vec{v}=(\nabla f(x,y)\cdot \vec{u})\vec{u}+(\nabla f(x,y)\cdot \vec{v})\vec{v}=((\frac{\partial f(x,y)}{\partial x},\frac{\partial f(x,y)}{\partial y})\cdot (u_1,u_2))(u_1,u_2) + ((\frac{\partial f(x,y)}{\partial x},\frac{\partial f(x,y)}{\partial y})\cdot (v_1,v_2))(v_1,v_2)=(\frac{\partial f(x,y)}{\partial x}u_1 + \frac{\partial f(x,y)}{\partial y}u_2)(u_1,u_2) + (\frac{\partial f(x,y)}{\partial x}v_1 + \frac{\partial f(x,y)}{\partial y}v_2)(v_1,v_2)= (\frac{\partial f(x,y)}{\partial x}u_1^2,\frac{\partial f(x,y)}{\partial x}u_1u_2) + (\frac{\partial f(x,y)}{\partial y}u_2u_1,\frac{\partial f(x,y)}{\partial y}u_2^2)+ (\frac{\partial f(x,y)}{\partial x}v_1^2,\frac{\partial f(x,y)}{\partial x}v_1v_2) + (\frac{\partial f(x,y)}{\partial y}v_2v_1,\frac{\partial f(x,y)}{\partial y}v_2^2)$

I don't know how to continue, am I doing well?
Can someone give me a suggestion? PLEASE!!
Thanks.

A:

The statement you wish to prove follows from the general fact that 
$$
w=w^\top uu + w^\top vv 
$$
for all $w$ whenever $u,v$ are orthonormal vectors in $R^2$. 
(Apply this to $w=\nabla f$. All you need to know about gradients to conclude is that $\partial_uf=\nabla f^\top u$)
The general fact is simply another way to express $x=QQ^\top x$, i.e. $QQ^\top=\text{Id}$ for orthonormal matrices.

