[MATH] Dérivée de...

Le 05 juin 2024 à 21:06:16 :
C'est pas compliqué khey, c'est la dérivée d'une fonction au carré multipliée par une constante.

Suffit d'appliquer la formule (u^2)'=... et tu dérives tout facilement !

ouais je sais mais je suis déjà tombé sur des trucs comme x^((1-x)/(x)) donc ça me fait hésiter pour tout vu que je comprend pas quand il faut remplacer

Tiens kheyou pour t'aider je te copie/colle une post de math.stackechange qui explique très bien la preuve de la règle de la chaine

Let f and g be differentiable and g non-constant. If g is the constant function, then clearly (fog)’ is zero since fog would be constant. So trivially it is true that (fog)’=f’(g(x))g’(x) since g’=0. I'll assume they're both defined on all of R
to avoid annoyances with domains and ranges.
Then we wish to evaluate (fog)’(x)
. From the definition of the derivative, we have
(fog)’(x)=ddxf(g(x))=limDELTAx->0f(g(x+DELTAx))-f(g(x))DELTAx.
Let's multiply by a clever form of 1
to make things easier on ourselves, particularly we'll multiply by
g(x+DELTAx)-g(x)g(x+DELTAx)-g(x).
This is where I required that g
be non-constant. If it were constant, the above expression would make no sense since we would be dividing 0 by 0. Note that this resembles the terms inside of f. This is not by accident. Additionally we have that g(x+DELTAx)~g(x)+g’(x)DELTAx for small DELTAx (this is what the derivative is for - linear approximations). If g’(x)=0, then g(x+DELTAx)~g(x)
and in this case, we have
limDELTAx->0f(g(x+DELTAx))-f(g(x))DELTAx=0.
Again, this is clearly equal to f’(g(x))g’(x)
since g’=0. If g’(x)=/=0, we can make use of our clever form of 1
to get
(fog)’(x)=limDELTAx->0f(g(x+DELTAx))-f(g(x))g(x+DELTAx)-g(x)g(x+DELTAx)-g(x)DELTAx.
Now both pieces look eerily like a derivative (which is what we want), except that the first piece has g
in it. However if DELTAx->0, we know that g(x+DELTAx)->g(x) since g is differentiable (and therefore continuous). Clearly limDELTAx->0g(x+DELTAx)-g(x)DELTAx exists since g is differentiable so we need only to argue that limDELTAx->0f(g(x+DELTAx)-f(g(x))g(x+DELTAx)-g(x)
is well-defined. By limit theorems we know that if both limits exists, we can distribute the limit to each piece and evaluate.
So we want to argue that
limDELTAx->0f(g(x+DELTAx))-f(g(x))g(x+DELTAx)-g(x)
is well-defined. Using our approximation for g(x+DELTAx)
from above, we have
limDELTAx->0f(g(x)+g’(x)DELTAx)-f(g(x))g(x)+g’(x)DELTAx-g(x).
Cancelling appropriate terms we have
limDELTAx->0f(g(x)+g’(x)DELTAx)-f(g(x))g’(x)DELTAx.
Repeating the same logic as above with f(g(x)+g’(x)DELTAx)
, we have that f(g(x)+g’(x)DELTAx)~f(g(x))+f’(g(x))g’(x)DELTAx
. And so we get
limDELTAx->0f(g(x))+f’(g(x))g’(x)DELTAx-f(g(x))g’(x)DELTAx=f’(g(x)).
Since the limit of the first piece makes sense and the limit of the second piece makes sense, we can distribute the limits to get that
(fog)’(x)=limDELTAx->0f(g(x+DELTAx))-f(g(x))g(x+DELTAx)-g(x)limDELTAx->0g(x+DELTAx)-g(x)DELTAx=f’(g(x))g’(x)
by our above calculations. So in each case that emerged we had that (fog)’(x)=f’(g(x))g’(x)
and so we conclude that the chain rule holds.

Le 05 juin 2024 à 21:15:51 :
Tiens kheyou pour t'aider je te copie/colle une post de math.stackechange qui explique très bien la preuve de la règle de la chaine

Let f and g be differentiable and g non-constant. If g is the constant function, then clearly (fog)’ is zero since fog would be constant. So trivially it is true that (fog)’=f’(g(x))g’(x) since g’=0. I'll assume they're both defined on all of R
to avoid annoyances with domains and ranges.
Then we wish to evaluate (fog)’(x)
. From the definition of the derivative, we have
(fog)’(x)=ddxf(g(x))=limDELTAx->0f(g(x+DELTAx))-f(g(x))DELTAx.
Let's multiply by a clever form of 1
to make things easier on ourselves, particularly we'll multiply by
g(x+DELTAx)-g(x)g(x+DELTAx)-g(x).
This is where I required that g
be non-constant. If it were constant, the above expression would make no sense since we would be dividing 0 by 0. Note that this resembles the terms inside of f. This is not by accident. Additionally we have that g(x+DELTAx)~g(x)+g’(x)DELTAx for small DELTAx (this is what the derivative is for - linear approximations). If g’(x)=0, then g(x+DELTAx)~g(x)
and in this case, we have
limDELTAx->0f(g(x+DELTAx))-f(g(x))DELTAx=0.
Again, this is clearly equal to f’(g(x))g’(x)
since g’=0. If g’(x)=/=0, we can make use of our clever form of 1
to get
(fog)’(x)=limDELTAx->0f(g(x+DELTAx))-f(g(x))g(x+DELTAx)-g(x)g(x+DELTAx)-g(x)DELTAx.
Now both pieces look eerily like a derivative (which is what we want), except that the first piece has g
in it. However if DELTAx->0, we know that g(x+DELTAx)->g(x) since g is differentiable (and therefore continuous). Clearly limDELTAx->0g(x+DELTAx)-g(x)DELTAx exists since g is differentiable so we need only to argue that limDELTAx->0f(g(x+DELTAx)-f(g(x))g(x+DELTAx)-g(x)
is well-defined. By limit theorems we know that if both limits exists, we can distribute the limit to each piece and evaluate.
So we want to argue that
limDELTAx->0f(g(x+DELTAx))-f(g(x))g(x+DELTAx)-g(x)
is well-defined. Using our approximation for g(x+DELTAx)
from above, we have
limDELTAx->0f(g(x)+g’(x)DELTAx)-f(g(x))g(x)+g’(x)DELTAx-g(x).
Cancelling appropriate terms we have
limDELTAx->0f(g(x)+g’(x)DELTAx)-f(g(x))g’(x)DELTAx.
Repeating the same logic as above with f(g(x)+g’(x)DELTAx)
, we have that f(g(x)+g’(x)DELTAx)~f(g(x))+f’(g(x))g’(x)DELTAx
. And so we get
limDELTAx->0f(g(x))+f’(g(x))g’(x)DELTAx-f(g(x))g’(x)DELTAx=f’(g(x)).
Since the limit of the first piece makes sense and the limit of the second piece makes sense, we can distribute the limits to get that
(fog)’(x)=limDELTAx->0f(g(x+DELTAx))-f(g(x))g(x+DELTAx)-g(x)limDELTAx->0g(x+DELTAx)-g(x)DELTAx=f’(g(x))g’(x)
by our above calculations. So in each case that emerged we had that (fog)’(x)=f’(g(x))g’(x)
and so we conclude that the chain rule holds.

merci khey je lirai ça demain