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I am going through J.W. Anderson's book $Hyperbolic$ $Geometry$, and I come along to the following statement made: "We note here that the concatenation of piecewise differentiable paths is again piecewise differentiable, while the concatenation of differentiable paths is not necessarily differentiable".

For the second part about differentiable paths, I understand it as I just need to take the paths $\overrightarrow{r}(x(t),y(t))$, $x(t) = t$, $y(t) = t$ and $\overrightarrow{r_1}(x_1(t),y_1(t))$, $x_1(t) = -t$, $y_1(t) = t$, $t\geq 0$ for both. Then there is a cusp when $t=0$ so the concatenation of the two paths is not differentiable. However, how can I prove the first statement that the concatenation of piecewise differentiable paths is again piecewise differentiable?

Ben

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    @Rasmus @ThomasRot Ok thanks guys very much for the clarifications.2011-04-18

1 Answers 1

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Well the only points at which the concatenation can fail to be differentiable, are the points where the curves that make up this curve are not differentiable, plus the endpoints of these curves. These are only a finite number. Hence the concatenation is piecewise differentiable.