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If $t=\ln(x)$, $y$ some function of $x$, and $\dfrac{dy}{dx}=e^{-t}\dfrac{dy}{dt}$, why would the second derivative of $y$ with respect to $x$ be: $$-e^{-t}\frac{dt}{dx}\frac {dy}{dt} + e^{-t}\frac{d^2y}{dt^2}\frac{dt}{dx}?$$

I know this links into the chain rule. I don't have a good intuition for why the first term has $\dfrac{dt}{dx}\dfrac{dy}{dt}$ (although I strongly feel it's such that we can change the variable, since this question arose in the context of a second order differential equation where $y$ was differentiated in terms of $x$, but the equation was non linear, so we had to make it linear by substitution). Moreover, the proper problem that I would plead to be adressed is why the second term is differentiated in the way that it is. Basically, my question is: why is the differential of $\dfrac{dt}{dx}\dfrac{dy}{dt}$ with respect to $x$ given as $\dfrac{d^2y}{dt^2}\dfrac{dt}{dx}$.

Preferable if english to explain any mathematical derivations, but any of your personal time to help out is always much appreciated.

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  • $\begingroup$ Please use Mathjax to format your math text. This link may be useful as a reference: meta.math.stackexchange.com/questions/5020/… $\endgroup$ – Dave Jun 7 '17 at 2:42
  • $\begingroup$ Please forgive me. I have an exam in 5 hours, filled with complex numbers, first order, second order differentials, polar coordinates, complex series summations... and a lot more. $\endgroup$ – Mathematician Jun 7 '17 at 2:54
  • $\begingroup$ It is often helpful to make a full substitution instead of using the same letters with invisible markers for implied meaning. Thus define $u(t)=y(e^t)$, $y(x)=u(\ln(x))$, so that the application of the chain rule becomes more natural. $\endgroup$ – LutzL Jun 7 '17 at 5:34
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With your substitution $t=\ln (x) $, the differentiation wrt $x $ is a differentiation wrt $t $ multiplied by $e^{-t} $. thus

$$\frac {d}{dx}(\frac {dy}{dx})=\frac {d}{dt}(\frac {dy}{dx})e^{-t} $$

$$=\frac {d}{dt}(\frac {dy}{dt}e^{-t})e^{-t} $$

$$=(\frac {d^2y}{dt^2}-\frac {dy}{dt})e^{-2t}$$

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\begin{align} \dfrac{dy}{dx}&=e^{-t}\dfrac{dy}{dt}\\ \implies\dfrac{d}{dx}\left(\dfrac{dy}{dx}\right)&=\dfrac{d}{dx}\left(e^{-t}\dfrac{dy}{dt}\right)\\ \implies\dfrac{d^2y}{dx^2}&=e^{-t}\cdot\dfrac{d}{dx}\left(\dfrac{dy}{dt}\right)+\dfrac{dy}{dt}\cdot\dfrac{d}{dx}\left(e^{-t}\right)\\ &=e^{-t}\cdot\dfrac{d}{dt}\left(\dfrac{dy}{dt}\right)\frac{dt}{dx}+\dfrac{dy}{dt}\cdot\frac{d}{dt}\left(e^{-t}\right)\frac{dt}{dx}\\ &=e^{-t}\cdot\frac{d^2y}{dt^2}\frac{dt}{dx}-e^{-t}\frac{dt}{dx}\frac{dy}{dt}\\ &=-e^{-t}\frac{dt}{dx}\frac{dy}{dt}+e^{-t}\frac{d^2y}{dt^2}\frac{dt}{dx} \end{align}

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