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I'm reading these notes but I don't know how is the chain rule being applied on page 92.

It says that we start with $\dfrac{dv}{du}=\dfrac{v'}{u'}$, having ruled out the case where $u'=0$. Then, he differentiates this wrt $u$ using the chain rule and goes this way

$$\dfrac{d^2v}{du^2}=\dfrac{d}{dv}\left(\dfrac{v'}{u'}\right)=\dfrac{u'v''-v'u''}{u'^2}\dfrac{1}{u'}$$

I don't really know why he got that. I understand $u$ and $v$ are functions of $t$, one does not depend on the other, so...

Thanks for any help.

Martin Sleziak
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user10586
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  • The derivative should be $\displaystyle \frac{d}{du} (v'/u')$, not $\displaystyle \frac{d}{dv} (v'/u')$ – user77528 May 12 '13 at 19:22
  • @user77528: This is better as a comment, although I realize you don't yet have enough reputation. You can certainly improve by describing the this is the quotient rule with the correction above. Regards – Amzoti May 12 '13 at 19:43

1 Answers1

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These are in fact my notes. You have a small typo: $\dfrac d{dv}$ should be $\dfrac d{du}$. But we're using the chain rule from single-variable calculus here: $$\frac{dw}{du}=\frac{\frac{dw}{dt}}{\frac{du}{dt}}\,.$$

Ted Shifrin
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