Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I was hoping that someone could explain to me the major steps involved in the following derivation, as I am fairly new to differential equations:


share|improve this question
It is most helpful (and strongly encouraged here) to post the equation found via your link directly in your post. I clicked on your link, and it certainly wouldn't take much effort to type & include the actual equation right in your post. It also would help if you could let us know what you've tried, or how you might approach this problem. That said, Welcome! –  amWhy Sep 2 '12 at 23:42

1 Answer 1

up vote 0 down vote accepted

There must be some extra information given beyond the quoted lines. Assuming that $$n=n(y)$$ $$y=y(z)$$ formal application of the product and chain rules gives: $$\frac{dn}{dz}\frac{dy}{dz}+n\frac{d^2 y}{dz^2}=\frac{dn}{dy}$$ $$\frac{dn}{dy}\frac{dy}{dz}\frac{dy}{dz}+n\frac{d^2 y}{dz^2}=\frac{dn}{dy}$$ $$\frac{dn}{dy}\left(\frac{dy}{dz}\right)^2+n\frac{d^2 y}{dz^2}=\frac{dn}{dy}$$ So unless $\frac{dy}{dz}\ne 0$ the given conclusion does not seem to hold.

share|improve this answer
Thank you for the answer. Are you sure, however, you don't mean that unless dy/dz = 0, the given conclusion does not seem to hold? –  John Roberts Sep 3 '12 at 2:37
Actually what you need is $$ \dfrac{dn}{dy} \left(\dfrac{dy}{dz} \right)^2 = 0$$ so either $dn/dy = 0$ or $dy/dz = 0$. –  Robert Israel Sep 3 '12 at 4:43

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.