Mixing
Help with a bit more rigorous proof of the chain rule.
$begingroup$
We need to show that $D_x(fcirc g)(a)=D_xf(g(a)) D_xg(a)$
$D_xf(g(a)) D_xg(a)=(lim_{yto g(a)} frac{f(y)-f(g(a))}{y-g(a)}) (lim_{xto a} frac{g(x)-(g(a)}{x-a}) $
Substituting $y=g(x)$
$D_xf(g(a)) D_xg(a)=(lim_{g(x)to g(a)} frac{f(g(x))-f(g(a))}{g(x)-g(a)}) (lim_{xto a} frac{g(x)-(g(a)}{x-a}) $
We know that $forall epsilon exists epsilon' $ such that
$0<|g(x)-g(a)|< epsilon' implies |frac{f(g(x))-f(g(a))}{g(x)-g(a)} - D_xf(g(a))| < epsilon$
We also know that $forall epsilon' exists delta $ such that
$0<|x-a|< delta implies |frac{g(x)-g(a)}{x-a}| < epsilon'$
Thus, we can conclude $forall epsilon exists delta $ such that
$0<|x-a|< delta implies |frac{f(g(x))-f(g(a))}{g(x)-g(a)}| < epsilon$
Thus,
$D_xf(g(a)) D_xg(a)=(lim_{xto a} frac{f(g(x))-f(g(a))}{g(x)-g(a)}) (lim_{xto a} frac{g(x)-(g(a)}{x-a}) $
Ignoring $g(x)=g(a)$ (For now) and applying the rule of product of limits, we get
$D_xf(g(a)) D_xg(a)=D_x(f circ g)(a)$
Here's where I need help. I can't find a way to get around the case where $g(x)=g(a)$. It would be really helpful if someone were to verify my proof so far as well.
real-analysis calculus proof-verification epsilon-delta
asked Jan 10 at 16:38
Star Platinum ZA WARUDOStar Platinum ZA WARUDO
33812
$endgroup$
|
show 2 more comments
$begingroup$
We need to show that $D_x(fcirc g)(a)=D_xf(g(a)) D_xg(a)$
$D_xf(g(a)) D_xg(a)=(lim_{yto g(a)} frac{f(y)-f(g(a))}{y-g(a)}) (lim_{xto a} frac{g(x)-(g(a)}{x-a}) $
Substituting $y=g(x)$
$D_xf(g(a)) D_xg(a)=(lim_{g(x)to g(a)} frac{f(g(x))-f(g(a))}{g(x)-g(a)}) (lim_{xto a} frac{g(x)-(g(a)}{x-a}) $
We know that $forall epsilon exists epsilon' $ such that
$0<|g(x)-g(a)|< epsilon' implies |frac{f(g(x))-f(g(a))}{g(x)-g(a)} - D_xf(g(a))| < epsilon$
We also know that $forall epsilon' exists delta $ such that
$0<|x-a|< delta implies |frac{g(x)-g(a)}{x-a}| < epsilon'$
Thus, we can conclude $forall epsilon exists delta $ such that
$0<|x-a|< delta implies |frac{f(g(x))-f(g(a))}{g(x)-g(a)}| < epsilon$
Thus,
$D_xf(g(a)) D_xg(a)=(lim_{xto a} frac{f(g(x))-f(g(a))}{g(x)-g(a)}) (lim_{xto a} frac{g(x)-(g(a)}{x-a}) $
Ignoring $g(x)=g(a)$ (For now) and applying the rule of product of limits, we get
$D_xf(g(a)) D_xg(a)=D_x(f circ g)(a)$
Here's where I need help. I can't find a way to get around the case where $g(x)=g(a)$. It would be really helpful if someone were to verify my proof so far as well.
real-analysis calculus proof-verification epsilon-delta
asked Jan 10 at 16:38
Star Platinum ZA WARUDOStar Platinum ZA WARUDO
33812
$endgroup$
$begingroup$
That is a problem. I believe it is solved by assuming the functions g is analytic and applying the local isolation of zeros theorem.
$endgroup$
– Tom
Jan 10 at 22:12
$begingroup$
@Tom Does the rest of the proof look fine?
$endgroup$
– Star Platinum ZA WARUDO
Jan 10 at 22:26
$begingroup$
No. It starts to go wrong when you substitute $y=g(x)$ since $frac{1}{g(x)-g(a)}$ may not be well defined in a deleted neighborhood of $a$. If you assume $g'(a)$ is non-zero it is sufficient to get such a neighborhood though.
$endgroup$
– Tom
Jan 10 at 22:55
$begingroup$
@Tom I meant apart from when $g(x)=g(a)$, is the proof correct for all other cases?
$endgroup$
– Star Platinum ZA WARUDO
Jan 10 at 23:09
$begingroup$
No, that's what makes the proof not correct. It is a key problem that needs to be sorted in order for it to be a proof.
$endgroup$
– Tom
Jan 10 at 23:17
|
show 2 more comments
$begingroup$
We need to show that $D_x(fcirc g)(a)=D_xf(g(a)) D_xg(a)$
$D_xf(g(a)) D_xg(a)=(lim_{yto g(a)} frac{f(y)-f(g(a))}{y-g(a)}) (lim_{xto a} frac{g(x)-(g(a)}{x-a}) $
Substituting $y=g(x)$
$D_xf(g(a)) D_xg(a)=(lim_{g(x)to g(a)} frac{f(g(x))-f(g(a))}{g(x)-g(a)}) (lim_{xto a} frac{g(x)-(g(a)}{x-a}) $
We know that $forall epsilon exists epsilon' $ such that
$0<|g(x)-g(a)|< epsilon' implies |frac{f(g(x))-f(g(a))}{g(x)-g(a)} - D_xf(g(a))| < epsilon$
We also know that $forall epsilon' exists delta $ such that
$0<|x-a|< delta implies |frac{g(x)-g(a)}{x-a}| < epsilon'$
Thus, we can conclude $forall epsilon exists delta $ such that
$0<|x-a|< delta implies |frac{f(g(x))-f(g(a))}{g(x)-g(a)}| < epsilon$
Thus,
$D_xf(g(a)) D_xg(a)=(lim_{xto a} frac{f(g(x))-f(g(a))}{g(x)-g(a)}) (lim_{xto a} frac{g(x)-(g(a)}{x-a}) $
Ignoring $g(x)=g(a)$ (For now) and applying the rule of product of limits, we get
$D_xf(g(a)) D_xg(a)=D_x(f circ g)(a)$
Here's where I need help. I can't find a way to get around the case where $g(x)=g(a)$. It would be really helpful if someone were to verify my proof so far as well.
real-analysis calculus proof-verification epsilon-delta
asked Jan 10 at 16:38
Star Platinum ZA WARUDOStar Platinum ZA WARUDO
33812
$endgroup$
We need to show that $D_x(fcirc g)(a)=D_xf(g(a)) D_xg(a)$
$D_xf(g(a)) D_xg(a)=(lim_{yto g(a)} frac{f(y)-f(g(a))}{y-g(a)}) (lim_{xto a} frac{g(x)-(g(a)}{x-a}) $
Substituting $y=g(x)$
$D_xf(g(a)) D_xg(a)=(lim_{g(x)to g(a)} frac{f(g(x))-f(g(a))}{g(x)-g(a)}) (lim_{xto a} frac{g(x)-(g(a)}{x-a}) $
We know that $forall epsilon exists epsilon' $ such that
$0<|g(x)-g(a)|< epsilon' implies |frac{f(g(x))-f(g(a))}{g(x)-g(a)} - D_xf(g(a))| < epsilon$
We also know that $forall epsilon' exists delta $ such that
$0<|x-a|< delta implies |frac{g(x)-g(a)}{x-a}| < epsilon'$
Thus, we can conclude $forall epsilon exists delta $ such that
$0<|x-a|< delta implies |frac{f(g(x))-f(g(a))}{g(x)-g(a)}| < epsilon$
Thus,
$D_xf(g(a)) D_xg(a)=(lim_{xto a} frac{f(g(x))-f(g(a))}{g(x)-g(a)}) (lim_{xto a} frac{g(x)-(g(a)}{x-a}) $
Ignoring $g(x)=g(a)$ (For now) and applying the rule of product of limits, we get
$D_xf(g(a)) D_xg(a)=D_x(f circ g)(a)$
Here's where I need help. I can't find a way to get around the case where $g(x)=g(a)$. It would be really helpful if someone were to verify my proof so far as well.
real-analysis calculus proof-verification epsilon-delta
real-analysis calculus proof-verification epsilon-delta
asked Jan 10 at 16:38
Star Platinum ZA WARUDOStar Platinum ZA WARUDO
33812
asked Jan 10 at 16:38
Star Platinum ZA WARUDOStar Platinum ZA WARUDO
33812
asked Jan 10 at 16:38
Star Platinum ZA WARUDOStar Platinum ZA WARUDO
33812
asked Jan 10 at 16:38
Star Platinum ZA WARUDOStar Platinum ZA WARUDO
33812
asked Jan 10 at 16:38
Star Platinum ZA WARUDOStar Platinum ZA WARUDO
33812
33812
$begingroup$
That is a problem. I believe it is solved by assuming the functions g is analytic and applying the local isolation of zeros theorem.
$endgroup$
– Tom
Jan 10 at 22:12
$begingroup$
@Tom Does the rest of the proof look fine?
$endgroup$
– Star Platinum ZA WARUDO
Jan 10 at 22:26
$begingroup$
No. It starts to go wrong when you substitute $y=g(x)$ since $frac{1}{g(x)-g(a)}$ may not be well defined in a deleted neighborhood of $a$. If you assume $g'(a)$ is non-zero it is sufficient to get such a neighborhood though.
$endgroup$
– Tom
Jan 10 at 22:55
$begingroup$
@Tom I meant apart from when $g(x)=g(a)$, is the proof correct for all other cases?
$endgroup$
– Star Platinum ZA WARUDO
Jan 10 at 23:09
$begingroup$
No, that's what makes the proof not correct. It is a key problem that needs to be sorted in order for it to be a proof.
$endgroup$
– Tom
Jan 10 at 23:17
|
show 2 more comments
$begingroup$
That is a problem. I believe it is solved by assuming the functions g is analytic and applying the local isolation of zeros theorem.
$endgroup$
– Tom
Jan 10 at 22:12
$begingroup$
@Tom Does the rest of the proof look fine?
$endgroup$
– Star Platinum ZA WARUDO
Jan 10 at 22:26
$begingroup$
No. It starts to go wrong when you substitute $y=g(x)$ since $frac{1}{g(x)-g(a)}$ may not be well defined in a deleted neighborhood of $a$. If you assume $g'(a)$ is non-zero it is sufficient to get such a neighborhood though.
$endgroup$
– Tom
Jan 10 at 22:55
$begingroup$
@Tom I meant apart from when $g(x)=g(a)$, is the proof correct for all other cases?
$endgroup$
– Star Platinum ZA WARUDO
Jan 10 at 23:09
$begingroup$
No, that's what makes the proof not correct. It is a key problem that needs to be sorted in order for it to be a proof.
$endgroup$
– Tom
Jan 10 at 23:17
$begingroup$
That is a problem. I believe it is solved by assuming the functions g is analytic and applying the local isolation of zeros theorem.
$endgroup$
– Tom
Jan 10 at 22:12
$begingroup$
That is a problem. I believe it is solved by assuming the functions g is analytic and applying the local isolation of zeros theorem.
$endgroup$
– Tom
Jan 10 at 22:12
$begingroup$
@Tom Does the rest of the proof look fine?
$endgroup$
– Star Platinum ZA WARUDO
Jan 10 at 22:26
$begingroup$
@Tom Does the rest of the proof look fine?
$endgroup$
– Star Platinum ZA WARUDO
Jan 10 at 22:26
$begingroup$
No. It starts to go wrong when you substitute $y=g(x)$ since $frac{1}{g(x)-g(a)}$ may not be well defined in a deleted neighborhood of $a$. If you assume $g'(a)$ is non-zero it is sufficient to get such a neighborhood though.
$endgroup$
– Tom
Jan 10 at 22:55
$begingroup$
No. It starts to go wrong when you substitute $y=g(x)$ since $frac{1}{g(x)-g(a)}$ may not be well defined in a deleted neighborhood of $a$. If you assume $g'(a)$ is non-zero it is sufficient to get such a neighborhood though.
$endgroup$
– Tom
Jan 10 at 22:55
$begingroup$
@Tom I meant apart from when $g(x)=g(a)$, is the proof correct for all other cases?
$endgroup$
– Star Platinum ZA WARUDO
Jan 10 at 23:09
$begingroup$
@Tom I meant apart from when $g(x)=g(a)$, is the proof correct for all other cases?
$endgroup$
– Star Platinum ZA WARUDO
Jan 10 at 23:09
$begingroup$
No, that's what makes the proof not correct. It is a key problem that needs to be sorted in order for it to be a proof.
$endgroup$
– Tom
Jan 10 at 23:17
$begingroup$
No, that's what makes the proof not correct. It is a key problem that needs to be sorted in order for it to be a proof.
$endgroup$
– Tom
Jan 10 at 23:17
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
$newcommand{R}{mathbb{R}}newcommanddx{frac{d}{dx}}$
$newcommandvara{a}$
Recall
For a function $f$ differentiable at $b$, we may write a linear approximation for $f$ in a neighborhood of $b$:
begin{equation}
newcommand{linapprox}[2]{f'({#1})({#2}-{#1}) + f({#1}) + o({#2})}
f(x)= linapprox{vara}{x} qquad(*)
end{equation}
Where $o(x)$ is the difference between $f'(vara)(x-vara) + f(vara)$ and $f(x)$. The differentiability of $f$ implies that $frac{o(x)-o(vara)}{x-vara}$ converges to $0$, and we will use this fact in a moment.
Let $f,g$ be continuous functions, $g$ differentiable at a point $a$, and $f$ differentiable at $g(a)$. Then, $fcirc g$ is also differentiable at $a$ and $(fcirc g)'(a) = (f'(g(a))g'(a)$
Proof
We write the definition of $(fcirc g)':$
begin{equation}
limlimits_{xrightarrow a}{frac{f(g(x))-f(g(a))}{x-a}}qquad(1)
end{equation}
Now, we will apply $(*)$ to get a linear approximation of $f$ in a neighborhood of $g(a)$ so that $(1)$ becomes the following:
begin{align}
&limlimits_{xrightarrow a}
frac{linapprox{g(a)}{g(x)}
-(linapprox{g(a)}{g(a)})}{x-a}\
&=
limlimits_{xrightarrow a}frac{
f'(g(a))(g(x)-g(a))+o(g(x)) +o(g(a))}{x-a}qquadtext{after cancelling like terms}
end{align}
In the above, $frac{o(x)-o(a)}{x-a}$ converges to $0$ as $x$ converges to $a$. Furthermore, $f'(g(a))$ is a constant, so we can move it outside of the limit.
begin{equation}
(fcirc g)'(a)= f'(g(a))
left[limlimits_{xrightarrow a}
frac{g(x)-g(a)}{x-a}
right]+
limlimits_{xrightarrow a}frac{o(x)-o(a)}{x-a}\\
=f'(g(a))g'(a)
end{equation}
Thus $(fcirc g)'(a)$ exists and is equal to $f'(g(a))g'(a)$
answered Jan 11 at 0:51
TomTom
736
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3068867%2fhelp-with-a-bit-more-rigorous-proof-of-the-chain-rule%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$newcommand{R}{mathbb{R}}newcommanddx{frac{d}{dx}}$
$newcommandvara{a}$
Recall
For a function $f$ differentiable at $b$, we may write a linear approximation for $f$ in a neighborhood of $b$:
begin{equation}
newcommand{linapprox}[2]{f'({#1})({#2}-{#1}) + f({#1}) + o({#2})}
f(x)= linapprox{vara}{x} qquad(*)
end{equation}
Where $o(x)$ is the difference between $f'(vara)(x-vara) + f(vara)$ and $f(x)$. The differentiability of $f$ implies that $frac{o(x)-o(vara)}{x-vara}$ converges to $0$, and we will use this fact in a moment.
Let $f,g$ be continuous functions, $g$ differentiable at a point $a$, and $f$ differentiable at $g(a)$. Then, $fcirc g$ is also differentiable at $a$ and $(fcirc g)'(a) = (f'(g(a))g'(a)$
Proof
We write the definition of $(fcirc g)':$
begin{equation}
limlimits_{xrightarrow a}{frac{f(g(x))-f(g(a))}{x-a}}qquad(1)
end{equation}
Now, we will apply $(*)$ to get a linear approximation of $f$ in a neighborhood of $g(a)$ so that $(1)$ becomes the following:
begin{align}
&limlimits_{xrightarrow a}
frac{linapprox{g(a)}{g(x)}
-(linapprox{g(a)}{g(a)})}{x-a}\
&=
limlimits_{xrightarrow a}frac{
f'(g(a))(g(x)-g(a))+o(g(x)) +o(g(a))}{x-a}qquadtext{after cancelling like terms}
end{align}
In the above, $frac{o(x)-o(a)}{x-a}$ converges to $0$ as $x$ converges to $a$. Furthermore, $f'(g(a))$ is a constant, so we can move it outside of the limit.
begin{equation}
(fcirc g)'(a)= f'(g(a))
left[limlimits_{xrightarrow a}
frac{g(x)-g(a)}{x-a}
right]+
limlimits_{xrightarrow a}frac{o(x)-o(a)}{x-a}\\
=f'(g(a))g'(a)
end{equation}
Thus $(fcirc g)'(a)$ exists and is equal to $f'(g(a))g'(a)$
answered Jan 11 at 0:51
TomTom
736
$endgroup$
add a comment |
$begingroup$
$newcommand{R}{mathbb{R}}newcommanddx{frac{d}{dx}}$
$newcommandvara{a}$
Recall
For a function $f$ differentiable at $b$, we may write a linear approximation for $f$ in a neighborhood of $b$:
begin{equation}
newcommand{linapprox}[2]{f'({#1})({#2}-{#1}) + f({#1}) + o({#2})}
f(x)= linapprox{vara}{x} qquad(*)
end{equation}
Where $o(x)$ is the difference between $f'(vara)(x-vara) + f(vara)$ and $f(x)$. The differentiability of $f$ implies that $frac{o(x)-o(vara)}{x-vara}$ converges to $0$, and we will use this fact in a moment.
Let $f,g$ be continuous functions, $g$ differentiable at a point $a$, and $f$ differentiable at $g(a)$. Then, $fcirc g$ is also differentiable at $a$ and $(fcirc g)'(a) = (f'(g(a))g'(a)$
Proof
We write the definition of $(fcirc g)':$
begin{equation}
limlimits_{xrightarrow a}{frac{f(g(x))-f(g(a))}{x-a}}qquad(1)
end{equation}
Now, we will apply $(*)$ to get a linear approximation of $f$ in a neighborhood of $g(a)$ so that $(1)$ becomes the following:
begin{align}
&limlimits_{xrightarrow a}
frac{linapprox{g(a)}{g(x)}
-(linapprox{g(a)}{g(a)})}{x-a}\
&=
limlimits_{xrightarrow a}frac{
f'(g(a))(g(x)-g(a))+o(g(x)) +o(g(a))}{x-a}qquadtext{after cancelling like terms}
end{align}
In the above, $frac{o(x)-o(a)}{x-a}$ converges to $0$ as $x$ converges to $a$. Furthermore, $f'(g(a))$ is a constant, so we can move it outside of the limit.
begin{equation}
(fcirc g)'(a)= f'(g(a))
left[limlimits_{xrightarrow a}
frac{g(x)-g(a)}{x-a}
right]+
limlimits_{xrightarrow a}frac{o(x)-o(a)}{x-a}\\
=f'(g(a))g'(a)
end{equation}
Thus $(fcirc g)'(a)$ exists and is equal to $f'(g(a))g'(a)$
answered Jan 11 at 0:51
TomTom
736
$endgroup$
add a comment |
$begingroup$
$newcommand{R}{mathbb{R}}newcommanddx{frac{d}{dx}}$
$newcommandvara{a}$
Recall
For a function $f$ differentiable at $b$, we may write a linear approximation for $f$ in a neighborhood of $b$:
begin{equation}
newcommand{linapprox}[2]{f'({#1})({#2}-{#1}) + f({#1}) + o({#2})}
f(x)= linapprox{vara}{x} qquad(*)
end{equation}
Where $o(x)$ is the difference between $f'(vara)(x-vara) + f(vara)$ and $f(x)$. The differentiability of $f$ implies that $frac{o(x)-o(vara)}{x-vara}$ converges to $0$, and we will use this fact in a moment.
Let $f,g$ be continuous functions, $g$ differentiable at a point $a$, and $f$ differentiable at $g(a)$. Then, $fcirc g$ is also differentiable at $a$ and $(fcirc g)'(a) = (f'(g(a))g'(a)$
Proof
We write the definition of $(fcirc g)':$
begin{equation}
limlimits_{xrightarrow a}{frac{f(g(x))-f(g(a))}{x-a}}qquad(1)
end{equation}
Now, we will apply $(*)$ to get a linear approximation of $f$ in a neighborhood of $g(a)$ so that $(1)$ becomes the following:
begin{align}
&limlimits_{xrightarrow a}
frac{linapprox{g(a)}{g(x)}
-(linapprox{g(a)}{g(a)})}{x-a}\
&=
limlimits_{xrightarrow a}frac{
f'(g(a))(g(x)-g(a))+o(g(x)) +o(g(a))}{x-a}qquadtext{after cancelling like terms}
end{align}
In the above, $frac{o(x)-o(a)}{x-a}$ converges to $0$ as $x$ converges to $a$. Furthermore, $f'(g(a))$ is a constant, so we can move it outside of the limit.
begin{equation}
(fcirc g)'(a)= f'(g(a))
left[limlimits_{xrightarrow a}
frac{g(x)-g(a)}{x-a}
right]+
limlimits_{xrightarrow a}frac{o(x)-o(a)}{x-a}\\
=f'(g(a))g'(a)
end{equation}
Thus $(fcirc g)'(a)$ exists and is equal to $f'(g(a))g'(a)$
answered Jan 11 at 0:51
TomTom
736
$endgroup$
$newcommand{R}{mathbb{R}}newcommanddx{frac{d}{dx}}$
$newcommandvara{a}$
Recall
For a function $f$ differentiable at $b$, we may write a linear approximation for $f$ in a neighborhood of $b$:
begin{equation}
newcommand{linapprox}[2]{f'({#1})({#2}-{#1}) + f({#1}) + o({#2})}
f(x)= linapprox{vara}{x} qquad(*)
end{equation}
Where $o(x)$ is the difference between $f'(vara)(x-vara) + f(vara)$ and $f(x)$. The differentiability of $f$ implies that $frac{o(x)-o(vara)}{x-vara}$ converges to $0$, and we will use this fact in a moment.
Let $f,g$ be continuous functions, $g$ differentiable at a point $a$, and $f$ differentiable at $g(a)$. Then, $fcirc g$ is also differentiable at $a$ and $(fcirc g)'(a) = (f'(g(a))g'(a)$
Proof
We write the definition of $(fcirc g)':$
begin{equation}
limlimits_{xrightarrow a}{frac{f(g(x))-f(g(a))}{x-a}}qquad(1)
end{equation}
Now, we will apply $(*)$ to get a linear approximation of $f$ in a neighborhood of $g(a)$ so that $(1)$ becomes the following:
begin{align}
&limlimits_{xrightarrow a}
frac{linapprox{g(a)}{g(x)}
-(linapprox{g(a)}{g(a)})}{x-a}\
&=
limlimits_{xrightarrow a}frac{
f'(g(a))(g(x)-g(a))+o(g(x)) +o(g(a))}{x-a}qquadtext{after cancelling like terms}
end{align}
In the above, $frac{o(x)-o(a)}{x-a}$ converges to $0$ as $x$ converges to $a$. Furthermore, $f'(g(a))$ is a constant, so we can move it outside of the limit.
begin{equation}
(fcirc g)'(a)= f'(g(a))
left[limlimits_{xrightarrow a}
frac{g(x)-g(a)}{x-a}
right]+
limlimits_{xrightarrow a}frac{o(x)-o(a)}{x-a}\\
=f'(g(a))g'(a)
end{equation}
Thus $(fcirc g)'(a)$ exists and is equal to $f'(g(a))g'(a)$
answered Jan 11 at 0:51
TomTom
736
edited Jan 24 at 20:37
answered Jan 11 at 0:51
TomTom
736
answered Jan 11 at 0:51
TomTom
736
answered Jan 11 at 0:51
TomTom
736
736
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3068867%2fhelp-with-a-bit-more-rigorous-proof-of-the-chain-rule%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
That is a problem. I believe it is solved by assuming the functions g is analytic and applying the local isolation of zeros theorem.
$endgroup$
– Tom
Jan 10 at 22:12
$begingroup$
@Tom Does the rest of the proof look fine?
$endgroup$
– Star Platinum ZA WARUDO
Jan 10 at 22:26
$begingroup$
No. It starts to go wrong when you substitute $y=g(x)$ since $frac{1}{g(x)-g(a)}$ may not be well defined in a deleted neighborhood of $a$. If you assume $g'(a)$ is non-zero it is sufficient to get such a neighborhood though.
$endgroup$
– Tom
Jan 10 at 22:55
$begingroup$
@Tom I meant apart from when $g(x)=g(a)$, is the proof correct for all other cases?
$endgroup$
– Star Platinum ZA WARUDO
Jan 10 at 23:09
$begingroup$
No, that's what makes the proof not correct. It is a key problem that needs to be sorted in order for it to be a proof.
$endgroup$
– Tom
Jan 10 at 23:17