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Jun 22, 2020 at 13:47 comment added Atom First, one doesn't talk of derivative of $f(t)$. It is derivative of $f$. Second, (in sloppy language) the derivative of $f(-t)$ wrt $t$ evaluated at $t$ is minus derivative of $f(t)$ wrt $t$ evaluated at $-t$. You can see how bad language can easily confuse.
Jun 22, 2020 at 13:40 comment added mike stone How is replacing $t$ by $-t$ an abuse of notation? That the deriviatve of $f(-t)$ wrt $t$ is minus the derivative of $f(t)$ wrt $t$ s just the chain rule of elementary calculus.
Jun 22, 2020 at 13:37 comment added Atom @mikestone I don’t like abusing notation. (And that’s a curse!)
Jun 22, 2020 at 13:35 comment added mike stone I found the original statement by Susskind much easier to understand. What is wrong with it? He simply replaces $t$ by $-t$ in his equation.
Jun 22, 2020 at 13:24 history edited Qmechanic CC BY-SA 4.0
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Jun 22, 2020 at 13:09 history asked Atom CC BY-SA 4.0