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This question has no particular applications; it just came upon me as I was thinking about change of variables in various scenarios.

Let $$\frac{df(x)}{dx}=g(x)$$

Suppose we wish to make a change of variable from $x$ to $u=p(x)$, so we substitute $x$ with $p^{-1}(u)$:

$$ \frac{df(p^{-1}(u))}{dx}=g(p^{-1}(u))$$

The left hand side (LHS) is a derivative with respect to $x$, so it's still a function of $x$. I'm unsure how to proceed further.

Suppose we invoke the chain rule:

$$ \frac{df(p^{-1}(u))}{du} \frac{du}{dx}=g(p^{-1}(u))$$

But LHS is still a function of $x$, because $\frac{du}{dx}$ is a function of $x$, while RHS is a function of $u$. But in a change of variable, the goal is to express everything in terms of $u$.

Rigor would be desirable.

FreshAir
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  • We can of course view the derivative as a limit and use the change of variable theorems for limits (such as this http://math.stackexchange.com/questions/456029/change-of-variables-in-limits), but I'm wondering if are other ideas... – FreshAir Aug 19 '15 at 10:25
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    We have $\frac{ \textrm{d} u }{\textrm{d}x} = \frac{ \textrm{d} p(x) }{\textrm{d}x} = p'(x) = p'( p^{-1}(u))$, which is a function of $u$, provided $p'$ exists. – Hetebrij Aug 19 '15 at 10:29
  • Thanks @Hetebrij. But if we are going to evaluate $\frac{ \textrm{d} u }{\textrm{d}x}$ at $x=p^{−1}(u)$, we don't need to do any substitution for the LHS in the first place: $$\frac{ \textrm{d} f(x) }{\textrm{d}x} \bigg|_{x=p^{-1}(u)}$$ – FreshAir Aug 19 '15 at 15:36

1 Answers1

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Using Inverse function theorem:

$$\frac{du}{dx} = \frac{1}{\frac{dx}{du}} = \frac{1}{\frac{dp^{-1}(u)}{du}} = p'(p^{-1}(u)).$$

hvedrung
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  • Interesting. So we used the inverse function theorem twice. But as @Hetebrij commented, can't we simply substitute $x=p^{-1}(u)$ without using the inverse function theorem: $$\frac{ \textrm{d} u }{\textrm{d}x} = p'(x) = p'( p^{-1}(u))$$? – FreshAir Aug 19 '15 at 15:33
  • Yes, @Hetebrij approach is right. – hvedrung Aug 19 '15 at 15:36