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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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  • $\begingroup$ What does it mean to differentiate with respect to $1-x$??? $\endgroup$ – copper.hat Feb 7 '18 at 18:45
  • $\begingroup$ @copper.hat I guess he considers $f(g(x))$ and wants to calculate $f^\prime(g(x))$ $\endgroup$ – Thomas Feb 7 '18 at 18:49
  • $\begingroup$ @Thomas: I was hoping the OP might express what they want a little more clearly. $\endgroup$ – copper.hat Feb 7 '18 at 18:50
  • $\begingroup$ Ok sorry, maybe it doesn't even make sense to do that. I was thinking that if I had some function of x, I could let u = 1 - x, and take the derivative with respect to u. The purpose of my question is that I learned that the deposit multiplier is the derivative of deposit size with respect to the reserve requirement. But my expression is MUCH easier to differentiate with respect to u = (1 - rr), if that is a thing I can do. $\endgroup$ – John Smith Feb 7 '18 at 18:51
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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 $\frac{d(1-x)}{dx} = -1$.

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    $\begingroup$ That is gross. ${}$ $\endgroup$ – copper.hat Feb 7 '18 at 18:49
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    $\begingroup$ The proof is just fine. True, it uses that little trick there in the second expression, which gives some budding mathematicians the chills...but what the heck! That's the same that is done in differential equations in different cases... $\endgroup$ – DonAntonio Feb 7 '18 at 18:53
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    $\begingroup$ That's what I understood from "differentiating with respect to $1-x$" as I indicated in the beginning. If OP doesn't mean that, I can simply edit this post according to what OP wants. $\endgroup$ – ArsenBerk Feb 7 '18 at 18:55
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    $\begingroup$ What is the formal definition of the "symbol" $\frac{df}{dg}$? $\endgroup$ – Ixion Feb 7 '18 at 19:02
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    $\begingroup$ a trick with notations is not a proof. The chain rule underneath doesn't justify any abuse of notation. So, go and learn what IS $df$. Good luck. $\endgroup$ – Netchaiev Feb 7 '18 at 19:14
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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 $2(1-x)$.

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