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If $x$ and $y$ are (non-constant) functions of $z$, is it in general true that $\frac{dy}{dx} = \frac{\frac{dy}{dz}}{\frac{dx}{dz}}$?

If it's not true, can you please provide a counterexample, and if it's true, how would we prove that?

Thanks!

S11n
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    Chain rule: $\frac{dy}{dz} = \frac{dy}{dx}\frac{dx}{dz}$ – psl2Z Feb 17 '24 at 16:58
  • Do you want to say that it boils down to $\frac{dx}{dz} = \frac{1}{\frac{dz}{dx}}$? What if $x(z)$ is not invertible? – S11n Feb 17 '24 at 17:13
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    I think the chain rule answers all of your questions. – psl2Z Feb 17 '24 at 17:15
  • It answers some questions, but in this particular examples it's harder to use the chain rule than it would be to use the approach I asked in the question...so it's not clear to me how to connect the two – S11n Feb 17 '24 at 17:17
  • @S11n These questions about $dy/dx$ will never stop. How do you know that your lovely formula holds when $x$ is not invertible? – Kurt G. Feb 17 '24 at 18:33
  • @KurtG. As Marco answered, and I hinted in the first comment here, seems like that holds only for invertible segments of $x$. – S11n Feb 22 '24 at 08:15

1 Answers1

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Assume $x,y,z$ differentiable functions and $$ \frac{dx}{dz}(z(p)) \ne 0 $$ In a neightbourhood of $z(p)$ (otherwise the formula does not make sense, as you are dividing by $0$) $$ \frac{dy}{dx}(p)=\frac{dy}{dz}(z(p))\frac{dz}{dx}(p) $$ for the chain rule.

If $\frac{dx}{dz}$ has constant sign in a neightbourhood $U$ of $z(p)$ the derivative $x(z)$ is strictly monotone in $U$. So it is invertible in $U$ and the formula holds.

But if the derivative is continuos and change sign in all the neightbourhoods of $z(p)$ then the derivative in $z(p)$ is zero, by Bolzano theorem. And this is against the hypothesis.

More generally, by Darboux theorem if the derivative change sign in any interval containing $z(p)$ than it has a zero in any such interval. As such if $$ \lim_{h \to z(p)} \frac{dx}{dz}(h) $$ exists the derivative need to have constant sign (and so the function is invertible) in a neightbourhood of $z(p)$

Marco
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