Show that the equation for $\nabla F$ yields the same result for all Cartesian coordinate systems We know that any 2 Cartesian coordinate systems are related by a rotation and a translation:
$x'=a+x\cos\alpha-y\sin\alpha$
$y'=b+x\sin\alpha+y\cos\alpha$
Consider the equation
$\nabla F = \frac{\partial F}{\partial x}i + \frac{\partial F}{\partial y}j$
The goal is to show that this equation yields the same results under the coordinate change above.
My approach was to compute $\nabla F'$, where
$F'(x', y')=F(a+x\cos\alpha-y\sin\alpha, b+x\sin\alpha+y\cos\alpha)$
Computing the partial derivatives of $F'$, I got:
$\frac{\partial F'}{\partial x'} = \frac{\partial F(a+x\cos\alpha-y\sin\alpha, b+x\sin\alpha+y\cos\alpha)}{\partial x} = \frac{\partial F}{\partial x}\cos\alpha i + \frac{\partial F}{\partial y}\sin\alpha j$
$\frac{\partial F'}{\partial y'} = \frac{\partial F(a+x\cos\alpha-y\sin\alpha, b+x\sin\alpha+y\cos\alpha)}{\partial y} = -\frac{\partial F}{\partial x}\sin\alpha i + \frac{\partial F}{\partial y}\cos\alpha j$
Summing them, I got
$\nabla F' = \frac{\partial F}{\partial x}(-\sin\alpha + \cos\alpha)i + \frac{\partial F}{\partial y}(\cos\alpha+\sin\alpha)j$
The new basis $i', j'$, as derived from the original transformation, is:
$i'=\cos\alpha i + \sin\alpha j$
$j'=-\sin\alpha i + \cos\alpha j$
I was hoping I could replace the trigonometric sums in the expression for $\nabla F'$ by $i',j'$. They look similar, but it seems like I am missing something.
For reference, this is the Exercise 1.4 from the book Introduction to Tensor Analysis and the Calculus of Moving Surfaces.
 A: The derivatives must be taken w.r.t. the new coordinates $x'$, $y'$. First note that $\frac{\partial}{\partial x'}$ and $\frac{\partial}{\partial y'}$ are not equal to $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial y}$. Also note that your expression for $\frac{\partial F'}{\partial x'}$, etc. cannot be correct, because the left hand side is a number and the right hand side a vector.
By the chain rule we have
$$
\begin{align}
\frac{\partial}{\partial x} &= \frac{\partial x'}{\partial x}\frac{\partial}{\partial x'} + \frac{\partial y'}{\partial x}\frac{\partial}{\partial y'} =
\phantom{-}\cos\alpha\frac{\partial}{\partial x'} + \sin\alpha\frac{\partial}{\partial y'} \\
%
\frac{\partial}{\partial y} &= \frac{\partial x'}{\partial y}\frac{\partial}{\partial x'} + \frac{\partial y'}{\partial y}\frac{\partial}{\partial y'} =
-\sin\alpha\frac{\partial}{\partial x'} + \cos\alpha\frac{\partial}{\partial y'}.
\tag{1}
\end{align}
$$
So
$$
\begin{align*}
\frac{\partial}{\partial x'} &=
\cos\alpha\frac{\partial}{\partial x} - \sin\alpha\frac{\partial}{\partial y} \\
%
\frac{\partial}{\partial y'} &=
\sin\alpha\frac{\partial}{\partial x} + \cos\alpha\frac{\partial}{\partial y}.
\tag{2}
\end{align*} 
$$
Now we can calculate $\frac{\partial F'}{\partial x'}$; the calculation for $\frac{\partial F'}{\partial y'}$ is very similar.
$$
\begin{align*}
 \frac{\partial F'}{\partial x'} &= \cos\alpha \frac{\partial F'}{\partial x} - \sin \alpha \frac{\partial F'}{\partial y} \\
&=\cos\alpha \left( \cos\alpha \frac{\partial F}{\partial x} + \sin \alpha \frac{\partial F}{\partial y'}\right) - \sin \alpha\left(-\sin\alpha \frac{\partial F}{\partial x} +\cos \alpha \frac{\partial F}{\partial y'}\right)\\
&= (\cos^2\alpha + \sin^2\alpha) \frac{\partial F}{\partial x} = \frac{\partial F}{\partial x}.
\end{align*}
$$
