5 ms·

Since 2 = e ^ ln 2, use the identity: 2^x = (e^(ln 2))^x = e ^ (x * ln 2)

VLM 11y ago
Well, that's the specific example, but there exist a class of educational blocks such that the learner doesn't know where they're stuck, because if they could define it, the act of correctly defining the stuck step would inherently unstick it. So MOOC strategy works for some areas that don't have sticking points like that, but not so well where five minutes with a Socratic strategy tutor could save a lot of video re-watching time. Arguably, better video would point out likely sticking points. However, advanced math instructors got their position not by getting education degrees which theoretically vocationally train people to present like that, but got their position by being very successful when getting a PHD in an obscure and advanced part of the field. So if you're lucky, your higher math instructor might be a talented educator in addition to being a talented mathematician... but probably not, unfortunately. It would be interesting as a thought experiment, or maybe as a startup, to see a good (emphasis on good) K-12 math instructor with a strong background in educational teaching skills try to teach diffeqs or higher math in general. I suspect they'd be extremely good at it, although probably very slow.
- huherto 11y ago
  A lot of those questions are taken care on the forums. People tend to stumble on the same sticking points. But it part of learning to learn in this new medium.
chris_wot 11y ago
Yeah, I get that bit, it's the next part I always get confused by. I'm sure I'll get it in the end. :-)
- mathgeek 11y ago
  If you haven't yet, watch the Khan Academy videos on the chain rule. Then use the related practice questions until you've mastered them. That's how I got the concepts drilled into my head for almost all of Calculus I and II.
  - chris_wot 11y ago
    That's kind of where I was getting confused :( I'm having a read of Better Explained at the moment: http://betterexplained.com/articles/derivatives-product-power-chain/ http://betterexplained.com/articles/derivatives-product-powe...
  - chris_wot 11y ago
    FWIW, I found the proof here: http://kruel.co/math/chainrule.pdf http://kruel.co/math/chainrule.pdf I'd actually tried to understand it earlier, but I got stuck at the point it says: We now use these equations to rewrite f (g(x + h)). In particular, use the first equation to obtain f(g(x + h)) = f(g(x) + [g′(x) + v]h), and use the second equation applied to the right-hand-side with k = [g′(x) + v]h and y = g(x). It was how they got k that tricked me. But then I realise that actually the bit I was missing was that y=g(x), and so as: f(g(x + h)) = f(g(x) + [g′(x) + v]h) You can actually make this: f(y + k) = f(y + [g'(x) + v]h) Thus k = (g'(x) + v)h

VLM 11y ago

Well, that's the specific example, but there exist a class of educational blocks such that the learner doesn't know where they're stuck, because if they could define it, the act of correctly defining the stuck step would inherently unstick it. So MOOC strategy works for some areas that don't have sticking points like that, but not so well where five minutes with a Socratic strategy tutor could save a lot of video re-watching time. Arguably, better video would point out likely sticking points. However, advanced math instructors got their position not by getting education degrees which theoretically vocationally train people to present like that, but got their position by being very successful when getting a PHD in an obscure and advanced part of the field. So if you're lucky, your higher math instructor might be a talented educator in addition to being a talented mathematician... but probably not, unfortunately. It would be interesting as a thought experiment, or maybe as a startup, to see a good (emphasis on good) K-12 math instructor with a strong background in educational teaching skills try to teach diffeqs or higher math in general. I suspect they'd be extremely good at it, although probably very slow.

huherto 11y ago
A lot of those questions are taken care on the forums. People tend to stumble on the same sticking points. But it part of learning to learn in this new medium.

chris_wot 11y ago

Yeah, I get that bit, it's the next part I always get confused by. I'm sure I'll get it in the end. :-)

mathgeek 11y ago
If you haven't yet, watch the Khan Academy videos on the chain rule. Then use the related practice questions until you've mastered them. That's how I got the concepts drilled into my head for almost all of Calculus I and II.
- chris_wot 11y ago
  That's kind of where I was getting confused :( I'm having a read of Better Explained at the moment: http://betterexplained.com/articles/derivatives-product-power-chain/ http://betterexplained.com/articles/derivatives-product-powe...
- chris_wot 11y ago
  FWIW, I found the proof here: http://kruel.co/math/chainrule.pdf http://kruel.co/math/chainrule.pdf I'd actually tried to understand it earlier, but I got stuck at the point it says: We now use these equations to rewrite f (g(x + h)). In particular, use the first equation to obtain f(g(x + h)) = f(g(x) + [g′(x) + v]h), and use the second equation applied to the right-hand-side with k = [g′(x) + v]h and y = g(x). It was how they got k that tricked me. But then I realise that actually the bit I was missing was that y=g(x), and so as: f(g(x + h)) = f(g(x) + [g′(x) + v]h) You can actually make this: f(y + k) = f(y + [g'(x) + v]h) Thus k = (g'(x) + v)h