Consider an image with edges, which we can take a gradient of (i.e. by subtracting pixels). Suppose we have a point $(x,0)$ on an edge of the image. Now suppose we rotate this point by an angle counterclockwise $\alpha$ to yield a new point $(x\cos(\alpha), x \sin(\alpha))$. I'm trying to figure out how to prove that the magnitude of the gradients at the point before and after are equal.
Here's what I've tried: We can see that the magnitude of the gradient of the point before only has a $y$ component. Thus it's magnitude is the partial of the image $\dfrac{\partial f(x,y)}{\partial y}$. Now for the rotated image, we are looking for $\sqrt{({{\dfrac{\partial f(x\cos(\alpha), x\sin(\alpha))}{\partial x}}})^2 + {\dfrac{\partial f(x\cos(\alpha), x\sin(\alpha))}{\partial y}})^2}$. We can see that the second squared term under the square root goes away by the chain rule so the expression becomes $\cos(\alpha)\sin(\alpha)\dfrac{\partial f(x,y)}{\partial x}$
...which is clearly not what we want.
Can anyone tell me what I'm doing wrong?