Question about Variable Change used in Differentiation

[hokhani]

Homework Statement

variable change of a differentiation

Relevant Equations

$\frac{df(x)}{dx}$

Consider differentiation $\frac{df(x)}{dx}$. If we change the variable as $x=-u$ how does the differentiation change? Does it change as $\frac{df(-u)}{-du}$?

 

[PeroK]

Homework Statement: variable change of a differentiation
Relevant Equations: $\frac{df(x)}{dx}$

Consider differentiation $\frac{df(x)}{dx}$. If we change the variable as $x=-u$ how does the differentiation change? Does it change as $\frac{df(-u)}{-du}$?

You must consider what a change of variable means. In general, you have a function $x = g(u)$. That allows us to define a new function by composition:
$$h(u) = f(x) = f(g(u))$$We can now apply the chain rule to get:
$$h'(u) = f'(g(u))g'(u)$$In the case where $g(u) = -u$, we have $g'(u) = -1$ and:
$$h'(u) = -f'(-u)$$Note that many textbooks will use the same function symbol $f$ for both the original function and the composite function $h$ here. And you end up with $$f'(-x) = -f'(x)$$or $$\frac{df(-x)}{dx} = - \frac{df(x)}{dx}$$PS note that, strictly speaking, these last two are wrong - but they are very common, especially in physics textbooks.

 

[FactChecker]

You need to be careful, especially for even functions, with how you interpret your expression, $\frac{df(-u)}{-du}$. You have not indicated what you think this means as a function of $u$. I think that you should apply the Chain rule as @PeroK describes. That will help you to keep the calculations straight.

 

[hokhani]

You must consider what a change of variable means. In general, you have a function $x = g(u)$. That allows us to define a new function by composition:
$$h(u) = f(x) = f(g(u))$$We can now apply the chain rule to get:
$$h'(u) = f'(g(u))g'(u)$$In the case where $g(u) = -u$, we have $g'(u) = -1$ and:
$$h'(u) = -f'(-u)$$Note that many textbooks will use the same function symbol $f$ for both the original function and the composite function $h$ here. And you end up with $$f'(-x) = -f'(x)$$or $$\frac{df(-x)}{dx} = - \frac{df(x)}{dx}$$PS note that, strictly speaking, these last two are wrong - but they are very common, especially in physics textbooks.

Thanks, by $f'(-u)$ do you mean $\frac{df(-u)}{d(-u)}$ or $\frac{df(-u)}{d(u)}$?

 

[Gavran]

$f'(-u)$ means $\frac{df(-u)}{d(-u)}$.

 

[PeroK]

Thanks, by $f'(-u)$ do you mean $\frac{df(-u)}{d(-u)}$ or $\frac{df(-u)}{d(u)}$?

I don't really use that notation myself. Given a differentiable function $f$, $f' \equiv \frac {df}{dx}$ is a well-defined function. Where the second formulation entails the use of a dummy variable $x$. Both $f$ and $f'$ can be applied to any variable, so we can write:
$$f(u), f'(u) \equiv \frac{df}{dx}(u) \equiv \frac{df}{dx}\bigg |_u$$I avoid things like $\frac{df(-u)}{d(-u)}$, as they are ambiguous and confusing to me.

 

[PeroK]

Let's take an example and let $f(x) = \sin x$, so that $f'(x) \equiv \frac{df}{dx} = \cos x$. In this case:
$$f(-u) = \sin(-u), \text{and} \ f'(-u) = \cos(-u)$$However, if we define:
$$h(u) = \sin(-u)$$then$$h'(u) \equiv \frac{dh}{du} = -\cos(-u) = -f'(u)$$Everything else involves an abuse of notation to some extent - which I prefer to avoid.

 

[Gavran]

$f'(-u)$ means $\frac{df(-u)}{d(-u)}$.

... and $-f'(-u)$ means $\frac{df(-u)}{du}$.

 

[FactChecker]

IMO, $f'(-u)$ is well defined as $f'(x)$ evaluated at $x=-u$. I'm not familiar with an official definition of $df(-u)$ or $df(-u)/du$.