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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 co
by 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 agoThat'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 agoFWIW, 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