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Forums
Mathematics
Calculus
Easy derivative but with a pesky singularity
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[QUOTE="snoopies622, post: 6869613, member: 112312"] I understand. My question is, since [itex] s\dot s = x\dot x + y \dot y [/itex] won't tell us [itex] \dot s [/itex] when s=0, how can we arrive at an equivalent formula that does? Can we mathematically manipulate [itex] s\dot s = x\dot x + y \dot y [/itex] in some way to get there? When we divide both sides by s, then we have that singularity at s=0. But it seems to me that this is not an asymptotic situation, so that singularity should be removable. But I don't know how to do it. [/QUOTE]
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Forums
Mathematics
Calculus
Easy derivative but with a pesky singularity
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