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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?

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

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    $\begingroup$ Just a remark: This computation is relevant in the context of Lie symmetries of differential equations. See Examples 2.21, 2.26 and 2.29 in Peter Olver's Applications of Lie Groups to Differential Equations. $\endgroup$ – Hans Lundmark Aug 16 '19 at 16:29
  • $\begingroup$ @HansLundmark Thank you for the useful reference---by chance I've just been reading through some of Olver's lecture notes. $\endgroup$ – Travis Willse Aug 16 '19 at 16:48

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