Derivative/Integral of a rotated function. I seem to have lost my other account, so this one is new.  Anyways, my questions is:
Is there a general way to produce an expression for a rotated curve that can then be integrated or differentiated like the original?  I read through some of the other comments on other threads and I realize that, upon rotation, the resulting curve may not even be a function, but, if it is, how can one proceed? 
For example, let $f$ be the function $f(x)=x^2$, $x\ge 0$, and let $g$ be the rotation of $f$ $135^\circ$ about the origin.  If, say, $(1,1)$ is on the original curve, one can rotate to get:
$$ \begin{pmatrix}
    -\displaystyle \sqrt2/2 & -\sqrt2/2  \\
    \sqrt2/2 & -\sqrt2/2 
  \end{pmatrix} \begin{pmatrix}
  1 \\
  1
  \end{pmatrix}$$
Which gives a single coordinate as the rotated point.  But, suppose I wanted to to find the derivative of $g$ at the rotated point?  It isn't sufficient (is it?) to say:
$$ \begin{pmatrix}
    -\displaystyle \sqrt2/2 & -\sqrt2/2  \\
    \sqrt2/2 & -\sqrt2/2 
  \end{pmatrix} \begin{pmatrix}
  x \\
  x^2
  \end{pmatrix}$$
Use the resulting $y$ coordinate, $\frac{\sqrt2}{2}x-\frac{\sqrt2}2x^2$, find the derivative of this curve, and plug in the $x$ value of the rotated point.
What is the appropriate way to do this?
P.S. No idea why the matrices are not appearing side by side.  I used the same latex code I always do...
 A: If the graph of a differentiable function $f$ is rotated and the rotated curve is again the graph of a differentiable function $g$ (perhaps on a small domain) then the slope of the graph is increased by the angle of rotation $\theta.$ So the derivatives before and after rotation should satisfy
$$\arctan g'(x_1)=\theta + \arctan f'(x_0)$$
Also, the point immediately "below" your curve on the rotated $X$-axis satisfies $x_0-x_1=f(x_0)\tan\theta.$ Combine this to obtain
$$\arctan g'(x_0-f(x_0)\tan\theta)=\theta + \arctan f'(x_0)$$
Similarly, when integrating $g$ over an interval of the new independent variable $x_1,$ take into account that $dx_1=(1-f'(x_0)\tan\theta)dx_0:$
$$\int_{x_1=a}^bg(x_1)dx_1=\int_{x_0=a_0}^{b_0}f(x_0-f(x_0)\tan\theta)(1-f'(x_0)\tan\theta)dx_0$$
where $a_0$ and $b_0$ are solutions of $a_0-a=f(a_0)\tan\theta$ and $b_0-b=f(b_0)\tan\theta.$
A: After rotation,
$$x'=Cx-Sy(x),\\y'=Sx+Cy(x)$$ and you would like to express $y'$ as a function of $x'$, right ?
Then you have to invert the equation 
$$x'=Cx-Sy(x)=f(x),$$ i.e. solve it for $x$, and get
$$x=f^{-1}(x').$$
From there,
$$y'=Sf^{-1}(x')+Cy\left(f^{-1}(x')\right).$$
Quite often though, $f(x)$ isn't invertible, and when it is, quite often it is not invertible analytically.

The case of the straight line $y=Ax+B$  is easy.
$$x'=Cx-S(Ax+B)$$ gives
$$x=\frac{x'+SB}{C-SA},\\
y'=(S+CA)\frac{x'+SB}{C-SA}+B,$$ obviously another straight line.
The case of the parabola, $y(x)=Ax^2+Bx+D$ is tractable.
$$x'=Cx-S(Ax^2+Bx+D).$$
This is a quadratic equation which you solve for $x$ by the usual formula. As there can be two solutions, you will need to sacrifice one branch.

Actually this approach by means of explicit curve equations $y=f(x)$ is pretty impractical, and you should prefer the parametric approach,
$$x=f(t),\\y=g(t)$$
for which more flexible tools are available. But this is another story.
