What is incorrect about this statement and/or proof?
$begingroup$
If $f$ and $g$ are two functions of $x$ such that $f$ is differentiable with respect to $x$ at $h$, $g$ is continuous, and $h=g(a)$ for some $a$, then $f$ is differentiable with respect to $g$.
Proof:
We are given that $$lim _{xrightarrow h} frac{f(x)-f(h)}{x-h}=T$$ and $$lim _{xrightarrow a} g(x)=g(a)$$
Through this we can infer that for every $varepsilon$ there exists a $delta$ such that $|frac{f(x)-f(h)}{x-h}-T|<varepsilon$ when $0<|x−h|<delta$
Now we replace $x$ by $g(x)$ to get $displaystyle left|frac{f(g(x))-f(g(a))}{g(x)-g(a)}-Tright|<varepsilon$ when $0<|g(x)−g(a)|<delta$
This implies that $$lim _{g(x)rightarrow g(a)} frac{f(g(x))-f(g(a))}{g(x)-g(a)}=T$$
Now this is obviously untrue, but I can't seem to find where exactly I went wrong. I would really appreciate if someone could explain which step was incorrect.
real-analysis epsilon-delta
edited Jan 10 at 0:19
El borito
666216
asked Jan 9 at 22:48
Star Platinum ZA WARUDOStar Platinum ZA WARUDO
34412
$endgroup$
add a comment |
$begingroup$
If $f$ and $g$ are two functions of $x$ such that $f$ is differentiable with respect to $x$ at $h$, $g$ is continuous, and $h=g(a)$ for some $a$, then $f$ is differentiable with respect to $g$.
Proof:
We are given that $$lim _{xrightarrow h} frac{f(x)-f(h)}{x-h}=T$$ and $$lim _{xrightarrow a} g(x)=g(a)$$
Through this we can infer that for every $varepsilon$ there exists a $delta$ such that $|frac{f(x)-f(h)}{x-h}-T|<varepsilon$ when $0<|x−h|<delta$
Now we replace $x$ by $g(x)$ to get $displaystyle left|frac{f(g(x))-f(g(a))}{g(x)-g(a)}-Tright|<varepsilon$ when $0<|g(x)−g(a)|<delta$
This implies that $$lim _{g(x)rightarrow g(a)} frac{f(g(x))-f(g(a))}{g(x)-g(a)}=T$$
Now this is obviously untrue, but I can't seem to find where exactly I went wrong. I would really appreciate if someone could explain which step was incorrect.
real-analysis epsilon-delta
edited Jan 10 at 0:19
El borito
666216
asked Jan 9 at 22:48
Star Platinum ZA WARUDOStar Platinum ZA WARUDO
34412
$endgroup$
1
$begingroup$
How do you know that $0<|g(x)-g(a)|$ in general?
$endgroup$
– Robert Wolfe
Jan 9 at 22:55
$begingroup$
@RobertWolfe Thanks for pointing that out. It seems that it is incorrect. Just curious, is the statement at the beginning true?
$endgroup$
– Star Platinum ZA WARUDO
Jan 9 at 23:03
add a comment |
$begingroup$
If $f$ and $g$ are two functions of $x$ such that $f$ is differentiable with respect to $x$ at $h$, $g$ is continuous, and $h=g(a)$ for some $a$, then $f$ is differentiable with respect to $g$.
Proof:
We are given that $$lim _{xrightarrow h} frac{f(x)-f(h)}{x-h}=T$$ and $$lim _{xrightarrow a} g(x)=g(a)$$
Through this we can infer that for every $varepsilon$ there exists a $delta$ such that $|frac{f(x)-f(h)}{x-h}-T|<varepsilon$ when $0<|x−h|<delta$
Now we replace $x$ by $g(x)$ to get $displaystyle left|frac{f(g(x))-f(g(a))}{g(x)-g(a)}-Tright|<varepsilon$ when $0<|g(x)−g(a)|<delta$
This implies that $$lim _{g(x)rightarrow g(a)} frac{f(g(x))-f(g(a))}{g(x)-g(a)}=T$$
Now this is obviously untrue, but I can't seem to find where exactly I went wrong. I would really appreciate if someone could explain which step was incorrect.
real-analysis epsilon-delta
edited Jan 10 at 0:19
El borito
666216
asked Jan 9 at 22:48
Star Platinum ZA WARUDOStar Platinum ZA WARUDO
34412
$endgroup$
If $f$ and $g$ are two functions of $x$ such that $f$ is differentiable with respect to $x$ at $h$, $g$ is continuous, and $h=g(a)$ for some $a$, then $f$ is differentiable with respect to $g$.
Proof:
We are given that $$lim _{xrightarrow h} frac{f(x)-f(h)}{x-h}=T$$ and $$lim _{xrightarrow a} g(x)=g(a)$$
Through this we can infer that for every $varepsilon$ there exists a $delta$ such that $|frac{f(x)-f(h)}{x-h}-T|<varepsilon$ when $0<|x−h|<delta$
Now we replace $x$ by $g(x)$ to get $displaystyle left|frac{f(g(x))-f(g(a))}{g(x)-g(a)}-Tright|<varepsilon$ when $0<|g(x)−g(a)|<delta$
This implies that $$lim _{g(x)rightarrow g(a)} frac{f(g(x))-f(g(a))}{g(x)-g(a)}=T$$
Now this is obviously untrue, but I can't seem to find where exactly I went wrong. I would really appreciate if someone could explain which step was incorrect.
real-analysis epsilon-delta
real-analysis epsilon-delta
edited Jan 10 at 0:19
El borito
666216
asked Jan 9 at 22:48
Star Platinum ZA WARUDOStar Platinum ZA WARUDO
34412
edited Jan 10 at 0:19
El borito
666216
asked Jan 9 at 22:48
Star Platinum ZA WARUDOStar Platinum ZA WARUDO
34412
edited Jan 10 at 0:19
El borito
666216
edited Jan 10 at 0:19
El borito
666216
edited Jan 10 at 0:19
El borito
666216
666216
asked Jan 9 at 22:48
Star Platinum ZA WARUDOStar Platinum ZA WARUDO
34412
asked Jan 9 at 22:48
Star Platinum ZA WARUDOStar Platinum ZA WARUDO
34412
asked Jan 9 at 22:48
Star Platinum ZA WARUDOStar Platinum ZA WARUDO
34412
34412
1
$begingroup$
How do you know that $0<|g(x)-g(a)|$ in general?
$endgroup$
– Robert Wolfe
Jan 9 at 22:55
$begingroup$
@RobertWolfe Thanks for pointing that out. It seems that it is incorrect. Just curious, is the statement at the beginning true?
$endgroup$
– Star Platinum ZA WARUDO
Jan 9 at 23:03
add a comment |
1
$begingroup$
How do you know that $0<|g(x)-g(a)|$ in general?
$endgroup$
– Robert Wolfe
Jan 9 at 22:55
$begingroup$
@RobertWolfe Thanks for pointing that out. It seems that it is incorrect. Just curious, is the statement at the beginning true?
$endgroup$
– Star Platinum ZA WARUDO
Jan 9 at 23:03
1
1
$begingroup$
How do you know that $0<|g(x)-g(a)|$ in general?
$endgroup$
– Robert Wolfe
Jan 9 at 22:55
$begingroup$
How do you know that $0<|g(x)-g(a)|$ in general?
$endgroup$
– Robert Wolfe
Jan 9 at 22:55
$begingroup$
@RobertWolfe Thanks for pointing that out. It seems that it is incorrect. Just curious, is the statement at the beginning true?
$endgroup$
– Star Platinum ZA WARUDO
Jan 9 at 23:03
$begingroup$
@RobertWolfe Thanks for pointing that out. It seems that it is incorrect. Just curious, is the statement at the beginning true?
$endgroup$
– Star Platinum ZA WARUDO
Jan 9 at 23:03
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The statement "$f $is differentiable with respect to $g$" does not make sense because $ g$ is a function not an independent variable.
Also you are dividing by $ g(x)-g(a)$ which is not allowed unless it is not zero.
answered Jan 9 at 23:07
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
$begingroup$
So what do we mean when we take the derivative of a function with respect to another function while using chain rule? Is it somehow different to what I did?
$endgroup$
– Star Platinum ZA WARUDO
Jan 9 at 23:11
$begingroup$
If you are trying to prove the chain rule, note that the derivatives are with respect to independent variables not with respect to functions. Please look at a real analysis book and see how the theorem is stated and proved.
$endgroup$
– Mohammad Riazi-Kermani
Jan 9 at 23:17
$begingroup$
Isn't $frac{df(g(x))}{dx} = frac{df(g(x))}{dg(x)} frac{dg(x)}{dx}$? So, what is the independent variable in $frac{df(g(x))}{dg(x)}$?
$endgroup$
– Star Platinum ZA WARUDO
Jan 9 at 23:24
$begingroup$
It is simply f'(g(x)) that is you take derivative of f(x) at x and then you replace x with g(x). It is a substitution. For example if you take derivative of sin(3x+1), you take derivative of sin(x) which is cos (x) and replace x with 3x+1 to get cos(3x+1) and multiply it with 3 which is derivative of 3x+1.
$endgroup$
– Mohammad Riazi-Kermani
Jan 10 at 1:54
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The statement "$f $is differentiable with respect to $g$" does not make sense because $ g$ is a function not an independent variable.
Also you are dividing by $ g(x)-g(a)$ which is not allowed unless it is not zero.
answered Jan 9 at 23:07
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
$begingroup$
So what do we mean when we take the derivative of a function with respect to another function while using chain rule? Is it somehow different to what I did?
$endgroup$
– Star Platinum ZA WARUDO
Jan 9 at 23:11
$begingroup$
If you are trying to prove the chain rule, note that the derivatives are with respect to independent variables not with respect to functions. Please look at a real analysis book and see how the theorem is stated and proved.
$endgroup$
– Mohammad Riazi-Kermani
Jan 9 at 23:17
$begingroup$
Isn't $frac{df(g(x))}{dx} = frac{df(g(x))}{dg(x)} frac{dg(x)}{dx}$? So, what is the independent variable in $frac{df(g(x))}{dg(x)}$?
$endgroup$
– Star Platinum ZA WARUDO
Jan 9 at 23:24
$begingroup$
It is simply f'(g(x)) that is you take derivative of f(x) at x and then you replace x with g(x). It is a substitution. For example if you take derivative of sin(3x+1), you take derivative of sin(x) which is cos (x) and replace x with 3x+1 to get cos(3x+1) and multiply it with 3 which is derivative of 3x+1.
$endgroup$
– Mohammad Riazi-Kermani
Jan 10 at 1:54
add a comment |
$begingroup$
The statement "$f $is differentiable with respect to $g$" does not make sense because $ g$ is a function not an independent variable.
Also you are dividing by $ g(x)-g(a)$ which is not allowed unless it is not zero.
answered Jan 9 at 23:07
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
$begingroup$
So what do we mean when we take the derivative of a function with respect to another function while using chain rule? Is it somehow different to what I did?
$endgroup$
– Star Platinum ZA WARUDO
Jan 9 at 23:11
$begingroup$
If you are trying to prove the chain rule, note that the derivatives are with respect to independent variables not with respect to functions. Please look at a real analysis book and see how the theorem is stated and proved.
$endgroup$
– Mohammad Riazi-Kermani
Jan 9 at 23:17
$begingroup$
Isn't $frac{df(g(x))}{dx} = frac{df(g(x))}{dg(x)} frac{dg(x)}{dx}$? So, what is the independent variable in $frac{df(g(x))}{dg(x)}$?
$endgroup$
– Star Platinum ZA WARUDO
Jan 9 at 23:24
$begingroup$
It is simply f'(g(x)) that is you take derivative of f(x) at x and then you replace x with g(x). It is a substitution. For example if you take derivative of sin(3x+1), you take derivative of sin(x) which is cos (x) and replace x with 3x+1 to get cos(3x+1) and multiply it with 3 which is derivative of 3x+1.
$endgroup$
– Mohammad Riazi-Kermani
Jan 10 at 1:54
add a comment |
$begingroup$
The statement "$f $is differentiable with respect to $g$" does not make sense because $ g$ is a function not an independent variable.
Also you are dividing by $ g(x)-g(a)$ which is not allowed unless it is not zero.
answered Jan 9 at 23:07
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
The statement "$f $is differentiable with respect to $g$" does not make sense because $ g$ is a function not an independent variable.
Also you are dividing by $ g(x)-g(a)$ which is not allowed unless it is not zero.
answered Jan 9 at 23:07
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
edited Jan 9 at 23:09
answered Jan 9 at 23:07
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered Jan 9 at 23:07
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered Jan 9 at 23:07
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
41.6k42061
$begingroup$
So what do we mean when we take the derivative of a function with respect to another function while using chain rule? Is it somehow different to what I did?
$endgroup$
– Star Platinum ZA WARUDO
Jan 9 at 23:11
$begingroup$
If you are trying to prove the chain rule, note that the derivatives are with respect to independent variables not with respect to functions. Please look at a real analysis book and see how the theorem is stated and proved.
$endgroup$
– Mohammad Riazi-Kermani
Jan 9 at 23:17
$begingroup$
Isn't $frac{df(g(x))}{dx} = frac{df(g(x))}{dg(x)} frac{dg(x)}{dx}$? So, what is the independent variable in $frac{df(g(x))}{dg(x)}$?
$endgroup$
– Star Platinum ZA WARUDO
Jan 9 at 23:24
$begingroup$
It is simply f'(g(x)) that is you take derivative of f(x) at x and then you replace x with g(x). It is a substitution. For example if you take derivative of sin(3x+1), you take derivative of sin(x) which is cos (x) and replace x with 3x+1 to get cos(3x+1) and multiply it with 3 which is derivative of 3x+1.
$endgroup$
– Mohammad Riazi-Kermani
Jan 10 at 1:54
add a comment |
$begingroup$
So what do we mean when we take the derivative of a function with respect to another function while using chain rule? Is it somehow different to what I did?
$endgroup$
– Star Platinum ZA WARUDO
Jan 9 at 23:11
$begingroup$
If you are trying to prove the chain rule, note that the derivatives are with respect to independent variables not with respect to functions. Please look at a real analysis book and see how the theorem is stated and proved.
$endgroup$
– Mohammad Riazi-Kermani
Jan 9 at 23:17
$begingroup$
Isn't $frac{df(g(x))}{dx} = frac{df(g(x))}{dg(x)} frac{dg(x)}{dx}$? So, what is the independent variable in $frac{df(g(x))}{dg(x)}$?
$endgroup$
– Star Platinum ZA WARUDO
Jan 9 at 23:24
$begingroup$
It is simply f'(g(x)) that is you take derivative of f(x) at x and then you replace x with g(x). It is a substitution. For example if you take derivative of sin(3x+1), you take derivative of sin(x) which is cos (x) and replace x with 3x+1 to get cos(3x+1) and multiply it with 3 which is derivative of 3x+1.
$endgroup$
– Mohammad Riazi-Kermani
Jan 10 at 1:54
$begingroup$
So what do we mean when we take the derivative of a function with respect to another function while using chain rule? Is it somehow different to what I did?
$endgroup$
– Star Platinum ZA WARUDO
Jan 9 at 23:11
$begingroup$
So what do we mean when we take the derivative of a function with respect to another function while using chain rule? Is it somehow different to what I did?
$endgroup$
– Star Platinum ZA WARUDO
Jan 9 at 23:11
$begingroup$
If you are trying to prove the chain rule, note that the derivatives are with respect to independent variables not with respect to functions. Please look at a real analysis book and see how the theorem is stated and proved.
$endgroup$
– Mohammad Riazi-Kermani
Jan 9 at 23:17
$begingroup$
If you are trying to prove the chain rule, note that the derivatives are with respect to independent variables not with respect to functions. Please look at a real analysis book and see how the theorem is stated and proved.
$endgroup$
– Mohammad Riazi-Kermani
Jan 9 at 23:17
$begingroup$
Isn't $frac{df(g(x))}{dx} = frac{df(g(x))}{dg(x)} frac{dg(x)}{dx}$? So, what is the independent variable in $frac{df(g(x))}{dg(x)}$?
$endgroup$
– Star Platinum ZA WARUDO
Jan 9 at 23:24
$begingroup$
Isn't $frac{df(g(x))}{dx} = frac{df(g(x))}{dg(x)} frac{dg(x)}{dx}$? So, what is the independent variable in $frac{df(g(x))}{dg(x)}$?
$endgroup$
– Star Platinum ZA WARUDO
Jan 9 at 23:24
$begingroup$
It is simply f'(g(x)) that is you take derivative of f(x) at x and then you replace x with g(x). It is a substitution. For example if you take derivative of sin(3x+1), you take derivative of sin(x) which is cos (x) and replace x with 3x+1 to get cos(3x+1) and multiply it with 3 which is derivative of 3x+1.
$endgroup$
– Mohammad Riazi-Kermani
Jan 10 at 1:54
$begingroup$
It is simply f'(g(x)) that is you take derivative of f(x) at x and then you replace x with g(x). It is a substitution. For example if you take derivative of sin(3x+1), you take derivative of sin(x) which is cos (x) and replace x with 3x+1 to get cos(3x+1) and multiply it with 3 which is derivative of 3x+1.
$endgroup$
– Mohammad Riazi-Kermani
Jan 10 at 1:54
add a comment |
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1
$begingroup$
How do you know that $0<|g(x)-g(a)|$ in general?
$endgroup$
– Robert Wolfe
Jan 9 at 22:55
$begingroup$
@RobertWolfe Thanks for pointing that out. It seems that it is incorrect. Just curious, is the statement at the beginning true?
$endgroup$
– Star Platinum ZA WARUDO
Jan 9 at 23:03