# Differentiating with respect to $1 - x$

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.

• What does it mean to differentiate with respect to $1-x$??? Commented Feb 7, 2018 at 18:45
• @copper.hat I guess he considers $f(g(x))$ and wants to calculate $f^\prime(g(x))$ Commented Feb 7, 2018 at 18:49
• @Thomas: I was hoping the OP might express what they want a little more clearly. Commented Feb 7, 2018 at 18:50
• 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. Commented Feb 7, 2018 at 18:51

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 $\frac{d(1-x)}{dx} = -1$.

• That is gross. ${}$ Commented Feb 7, 2018 at 18:49
• 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... Commented Feb 7, 2018 at 18:53
• What is the formal definition of the "symbol" $\frac{df}{dg}$? Commented Feb 7, 2018 at 19:02
• 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. Commented Feb 7, 2018 at 19:14
• @ArsenBerk And that step is derivative of the inverse, which is also due to the chain rule. It is just that people remember being scared about Leibniz but often times don't remember exactly what is it what they are supposed to be scared about. Commented Feb 7, 2018 at 19:45

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 $2(1-x)$.

I agree with @ArsenBerk generally, but I think there is in fact an even easier way. If you view the differential itself as an operator, then the operation is even easier. Rather than thinking about derivatives, you can think of the differential $$d()$$ as being an operator in and of itself, and the derivative as simply being a ratio of differentials. So, for instance, $$d(x^2) = 2x\,dx$$ and $$d(xy) = x\,dy + y\,dx$$ and so forth. So, in this case, you have:

$$\frac{dy}{d(1 - x)}$$

The numerator is already simplified, we just have to apply the differential to the denominator. Using the addition rule, $$d(1 - x) = d(1) - d(x)$$. The differential of a constant is zero, and the differential of $$x$$ is just $$dx$$ by definition. Therefore, $$d(1 - x) = -dx$$. Therefore, you can very simply state:

$$\frac{dy}{d(1 - x)} = \frac{dy}{d(1) - d(x)} = \frac{dy}{0 - dx} = -\frac{dy}{dx}$$

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.

I think it's too late to go for it but... nah. Here is a simpler question that you might be asking:

Find $$\frac{d}{dt}(3t^2 + 3)$$ and $$\frac{d}{d(1-t)}(3t^2 + 3)$$ Too solve this question, we can use the rule of parametric differentiation. Simply put: we can write $$y$$ as $$y = 3x^2+3$$ and $$x$$ as $$x = t$$ & $$x=1-t$$. When performing the parametric differentiation: $$\frac{\frac{dy}{dt}(3t^2+3) }{\frac{dx}{dt} (t)}$$ Similarly: $$\frac{\frac{dy}{dt}(3t^2+3) }{\frac{dx}{dt} (1-t)}$$ Solving the derivative, we get (for $$x = t$$ and $$1-t$$ respectively): $$\frac{6t}{1} = 6t$$ and $$\frac{6t}{-1} = -6t$$ Hope this helps!