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I have been working on this for a while and it makes no sense to me. Here's the deal.

We're using the substitution $X = \dfrac{x}{\delta}.$ Then, for some reason, this gives us

$$\begin{split} X = \dfrac{x}{\delta} & \iff \dfrac{d}{dx} = \dfrac{1}{\delta} \dfrac{d}{dX} \\ & \iff \dfrac{d^2}{dx^2} = \dfrac{1}{\delta^2} \dfrac{d^2}{dX^2} \end{split}$$

This makes no sense to me. Why isn't it

$$\begin{split} X = \dfrac{x}{\delta} & \iff \dfrac{d}{dX} = \dfrac{1}{\delta} \dfrac{d}{dx} \\ & \iff \dfrac{d^2}{dX^2} = \dfrac{1}{\delta} \dfrac{d^2}{dx^2}. \end{split}$$

Because I was under the impression that the derivative of $X$ would be $\frac{d}{dX}$ and the derivative of $\dfrac{x}{\delta}$ would be $\dfrac{1}{\delta}\dfrac{d}{dx}$ and that since $\dfrac{1}{\delta}$ is a coefficient, taking the derivative just leaves it there (instead of getting the coefficient $\dfrac{1}{\delta^2}.$

The only thing that anyone every says to me is that it's the chain rule. Which doesn't make sense to me because I thought the chain rule was

$$\left(\begin{array}{c} \text{derivative of} \\ \text{the outside} \end{array} \right) \cdot \left( \begin{array}{c} \text{derivative of} \\ \text{the inside} \end{array} \right)$$

If that's the case, what is the outside function and what is the inside function?

NOTE: If anyone wants any more context than what I gave, we are using these to solve second order linear differential equations, specifically as it applies to boundary layers. If you want more info than this, I'd be happy to provide some.

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    $\begingroup$ I recommend picking a favorite function - say, $f(x)=\sin x$ - and differentiating twice with respect to $x$, then writing the function in terms of $X$ and differentiating twice with respect to $X$. $\endgroup$ Nov 11, 2014 at 22:26

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What this actually means is that $X(x) = \frac x\delta$ is to be composed with a function in $x, g(x)$ such that we obtain a function in $X, f(X)$, hence.

$$f(X(x)) = f(\frac x\delta)$$ and $$\frac {d^2}{dx^2} f(X(x)) = \frac{d^2}{dx^2} f(\frac x\delta) = \frac d{dx} \Big( f'(\frac x\delta) \cdot \underbrace{\frac d{dx} X(x)}_{=\frac1\delta} \Big) \\ = \frac1\delta \frac d{dx} f'(\frac x\delta) = \frac1\delta f''(\frac x\delta) \cdot \underbrace{\frac d{dx} X(x)}_{=\frac1\delta} = \frac1{\delta^2} f''(\frac x\delta) = \frac1{\delta^2} f''(X)$$ Where $f'' = \frac {d^2}{dX^2} f$, so we can say that $$\frac {d^2}{dx^2} = \frac1{\delta^2} \frac{d^2}{dX^2}$$ This is all a bit unclean because you think of $\frac d{dX}$ as a function where it is in fact an operator (It operates on functions).

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  • $\begingroup$ How, then, are you getting $\dfrac{1}{\delta^2}$ out of the deal? $\endgroup$
    – Bark Jr.
    Nov 11, 2014 at 22:33
  • $\begingroup$ @BarkJr. Has the application of the chain rule become clearer now? $\endgroup$
    – AlexR
    Nov 11, 2014 at 22:40
  • $\begingroup$ Yes it has. Thank you for that thorough edit. $\endgroup$
    – Bark Jr.
    Nov 11, 2014 at 22:49
  • $\begingroup$ @BarkJr. Happy to help :) If anything remained unclear, do not hesitate to ask! $\endgroup$
    – AlexR
    Nov 11, 2014 at 22:56

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