I am fairly sure this is a silly question, but a Google search was insufficient to find a satisfactory answer.
If I differentiate some function of $x$ with respect to $1-x$, what do I get compared to differentiating with respect to $x$?
I know I need to use the chain rule to figure this out, but I am stuck on the details.
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$\begingroup$If you mean $\frac{dy}{d(1-x)}$, that is $$\frac{dy}{d(1-x)} = \frac{dy}{dx} \cdot \frac{dx}{d(1-x)} = \frac{\frac{dy}{dx}}{\frac{d(1-x)}{dx}} = -\frac{dy}{dx}$$ because $\frac{d(1-x)}{dx} = -1$.
$\endgroup$ 19 $\begingroup$If you have a function of $1-x$ and you want to differentiate that function with respect to $1-x$, start by setting $y:= 1 - x$. Then replace every $1-x$ with $y$ in the expression, and differentiate the expression with respect to the variable $y$. After differentiating, replace each of the $y$'s in the derivative with $1-x$.
For example, to differentiate $(1 - x)^{2}$ with respect to $1-x$, set $y := 1 - x$, and so the expression is now $y^{2}$. Differentiating this with respect to $y$ gives $2y$, and substituting $1-x$ back in gives the derivative as $2(1-x)$.
$\endgroup$ $\begingroup$Too long for a comment: I wanted to add that the manipulation in the answer of @ArsenBerk can be justified formally using the notion of differentiation respect to a function of limited variation of Daniell.
Also we can read the article of Pouso and Rodriguez A New Unification of Continuous, Discrete, and Impulsive Calculus through Stieltjes Derivatives for an amazing generalization of this idea.
Following the last cited article we would have something like: let $f,g:\Bbb R \to \Bbb R $ and $g$ monotone and left (or right) continuous, then we can define the derivative of $f$ at $x$ with respect to $g$ as$$ f'_g(x):=\begin{cases} \lim_{t\to x}\frac{f(t)-f(x)}{g(t)-g(x)},\quad \text{ if }g\text{ is continuous at }x\\ \lim_{t\to x^+}\frac{f(t)-f(x)}{g(t)-g(x)},\quad \text{ otherwise } \end{cases} $$if the corresponding limit exists (the lateral limit is the other if we choose $g$ right continuous instead of left continuous).
This $g$-derivative defines the Radon-Nikodym derivative of a Lebesgue-Stieltjes measure defined by $f$ (when this is possible) respect to the Lebesgue-Stieltjes measure defined by $g$ (when $df\ll dg$, of course), and a more general result can be found in the context of the Kurzweil-Stieltjes integration.
Therefore when $f$ and $g$ are normally differentiable at $x$, and $g'(x)\neq 0$, we have that $f'_g(x)=f'(x)/g'(x)$, what is the same result obtained by @ArsenBerk and his "heuristic" (maybe esoteric?) manipulation.
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