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.
On
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 $2(1-x)$.
On
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.
On
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!
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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:
$$\frac{dy}{d(1 - x)} = \frac{dy}{d(1) - d(x)} = \frac{dy}{0 - dx} = -\frac{dy}{dx}$$
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 $\frac{d(1-x)}{dx} = -1$.