# How is the derivative of a function $f$ related to the derivative of a function whose graph is the image of the graph of $f$ under a rotation?

The aim of this self-answered question is to record an answer worked out to a question posted earlier today that was deleted before the answer could be submitted.

Suppose $$f : U \to \Bbb R$$, $$U \subseteq \Bbb R$$, is a function differentiable at a point $$x$$. If the image of the graph of $$f$$ under a rotation $$R$$ is the graph of some function $$\tilde f$$, is $$\tilde f$$ differentiable at $$\tilde x := p(R(x, f(x)))$$, where $$p$$ is the projection onto the first coordinate?

• Happy you have done it, because it's a pity that the OP has erased it. – Jean Marie Aug 15 '19 at 22:16
• Should the question be community wiki, then? – Bladewood Aug 16 '19 at 15:46

By translation we may as well assume the rotation is about the origin, say, by anticlockwise angle $$\alpha$$. Then, for any $$x \in U$$ we have $$(\tilde x, \tilde f (\tilde x)) = (x \cos \alpha - f(x) \sin \alpha, x \sin \alpha + f(x) \cos \alpha),$$ and some algebra gives that we can rewrite $$Q := \frac{\tilde f(\widetilde{x + \Delta x}) - \tilde f(\tilde x)}{\widetilde{x + \Delta x} - \tilde x}$$ as $$Q = \frac{D + \tan \alpha}{1 - D \tan \alpha},$$ where $$D$$ is the difference quotient $$\frac{f(x + \Delta x) - f(x)}{\Delta x} .$$ Taking the limit as $$\Delta x \to 0$$ gives that $$\tilde f$$ is differentiable at $$\tilde x$$, provided $$f'(x) \neq \cot \alpha$$: $${\tilde f}'(\tilde x) = \lim_{\Delta x \to 0} Q = \lim_{\Delta x \to 0} \frac{D + \tan \alpha}{1 - D \tan \alpha} = \frac{f'(x) + \tan \alpha}{1 - f'(x) \tan \alpha} .$$ Here, the first equality follows from the fact that $$\widetilde{x + \Delta x} - \tilde x \to 0$$ as $$\Delta x \to 0$$ (which again uses that $$f$$ is differentiable at $$x$$) and the last equality is just uses continuity and the definition $$f'(x) = \lim_{\Delta x \to 0} D$$.
We can rewrite that formula for $$\alpha$$ in a particular range as $$\arctan \tilde f'(\tilde x) - \arctan f'(x) = \alpha , \qquad \alpha \in \left(-\frac{\pi}{2} - \arctan f'(x), \frac{\pi}{2} - \arctan f'(x)\right) ,$$ which we can interpret as the expected result that the angle between the tangent lines to the original and rotated functions is $$\alpha$$.
If $$f'(x) = \cot \alpha$$, then $$Q$$ is undefined, and rotating the graph of $$f$$ by an angle $$\alpha$$ maps the tangent line to $$f$$ at $$x$$ to a vertical line. We can view the restriction on $$\alpha$$ in the previous display equation as the requirement that the rotation does not rotate the tangent line to $$f$$ at $$x$$ beyond the vertical. If we relax the restriction on $$\alpha$$ to the condition $$f'(x) \neq \cot \alpha$$ we still have $$\arctan \tilde f'(\tilde x) - \arctan f'(x) \equiv \alpha \pmod \pi .$$
If we are happy to restrict to $$C^1$$ functions, we can quickly guarantee the existence of the derivative $$\tilde f'(\tilde x)$$ (again, with the restriction $$f'(x) \neq \cot \alpha$$) using the Implicit Function Theorem.