Can we apply the following derivatives to $f(y_1)$ depending on Leibniz Rule?
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Can we apply the following derivatives to $f(y_1)$ as below depending on Leibniz Rule?
$$f(y)= lim_{uto infty}{frac{g(y)-int_{1}^{u}{frac{x-lfloor x rfloor}{x^{1+y}}dx}}{h(y)-int_{1}^{u}{frac{x-lfloor x rfloor}{x^{2-y}}dx}}}$$
Now let us supose that the limit results of $f(y_1)$ will look as $0/0$. Thus by this advantage, can we apply L'Hopital rule to the right side with respect to the parameter $u$?
$$f(y_1)= lim_{uto infty}{frac{{frac {d} {du}}(g(y_1)-int_{1}^{u}{frac{x-lfloor x rfloor}{x^{1+y_1}}dx)}}{{frac {d} {du}}(h(y_1)-int_{1}^{u}{frac{x-lfloor x rfloor}{x^{2-y_1}}dx)}}}$$
$$f(y_1)= lim_{uto infty}{frac{{frac{u-lfloor u rfloor}{u^{1+y_1}}}}{{frac{u-lfloor u rfloor}{u^{2-y_1}}}}}$$
$$ f(y_1)=lim_{uto infty}frac{u^{2-y_1}}{u^{1+y_1}}$$
Is my way correct ?
real-analysis calculus functions
asked Dec 4 at 16:09
Testform
14
add a comment |
up vote
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Can we apply the following derivatives to $f(y_1)$ as below depending on Leibniz Rule?
$$f(y)= lim_{uto infty}{frac{g(y)-int_{1}^{u}{frac{x-lfloor x rfloor}{x^{1+y}}dx}}{h(y)-int_{1}^{u}{frac{x-lfloor x rfloor}{x^{2-y}}dx}}}$$
Now let us supose that the limit results of $f(y_1)$ will look as $0/0$. Thus by this advantage, can we apply L'Hopital rule to the right side with respect to the parameter $u$?
$$f(y_1)= lim_{uto infty}{frac{{frac {d} {du}}(g(y_1)-int_{1}^{u}{frac{x-lfloor x rfloor}{x^{1+y_1}}dx)}}{{frac {d} {du}}(h(y_1)-int_{1}^{u}{frac{x-lfloor x rfloor}{x^{2-y_1}}dx)}}}$$
$$f(y_1)= lim_{uto infty}{frac{{frac{u-lfloor u rfloor}{u^{1+y_1}}}}{{frac{u-lfloor u rfloor}{u^{2-y_1}}}}}$$
$$ f(y_1)=lim_{uto infty}frac{u^{2-y_1}}{u^{1+y_1}}$$
Is my way correct ?
real-analysis calculus functions
asked Dec 4 at 16:09
Testform
14
The question keeps changing and I cannot continue to update my answer. Please ask another question instead of changing it
– Federico
Dec 4 at 17:37
@Federico , Sorry, but this last shape, if I can take your comment, I will be glad.
– Testform
Dec 4 at 17:49
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Can we apply the following derivatives to $f(y_1)$ as below depending on Leibniz Rule?
$$f(y)= lim_{uto infty}{frac{g(y)-int_{1}^{u}{frac{x-lfloor x rfloor}{x^{1+y}}dx}}{h(y)-int_{1}^{u}{frac{x-lfloor x rfloor}{x^{2-y}}dx}}}$$
Now let us supose that the limit results of $f(y_1)$ will look as $0/0$. Thus by this advantage, can we apply L'Hopital rule to the right side with respect to the parameter $u$?
$$f(y_1)= lim_{uto infty}{frac{{frac {d} {du}}(g(y_1)-int_{1}^{u}{frac{x-lfloor x rfloor}{x^{1+y_1}}dx)}}{{frac {d} {du}}(h(y_1)-int_{1}^{u}{frac{x-lfloor x rfloor}{x^{2-y_1}}dx)}}}$$
$$f(y_1)= lim_{uto infty}{frac{{frac{u-lfloor u rfloor}{u^{1+y_1}}}}{{frac{u-lfloor u rfloor}{u^{2-y_1}}}}}$$
$$ f(y_1)=lim_{uto infty}frac{u^{2-y_1}}{u^{1+y_1}}$$
Is my way correct ?
real-analysis calculus functions
asked Dec 4 at 16:09
Testform
14
Can we apply the following derivatives to $f(y_1)$ as below depending on Leibniz Rule?
$$f(y)= lim_{uto infty}{frac{g(y)-int_{1}^{u}{frac{x-lfloor x rfloor}{x^{1+y}}dx}}{h(y)-int_{1}^{u}{frac{x-lfloor x rfloor}{x^{2-y}}dx}}}$$
Now let us supose that the limit results of $f(y_1)$ will look as $0/0$. Thus by this advantage, can we apply L'Hopital rule to the right side with respect to the parameter $u$?
$$f(y_1)= lim_{uto infty}{frac{{frac {d} {du}}(g(y_1)-int_{1}^{u}{frac{x-lfloor x rfloor}{x^{1+y_1}}dx)}}{{frac {d} {du}}(h(y_1)-int_{1}^{u}{frac{x-lfloor x rfloor}{x^{2-y_1}}dx)}}}$$
$$f(y_1)= lim_{uto infty}{frac{{frac{u-lfloor u rfloor}{u^{1+y_1}}}}{{frac{u-lfloor u rfloor}{u^{2-y_1}}}}}$$
$$ f(y_1)=lim_{uto infty}frac{u^{2-y_1}}{u^{1+y_1}}$$
Is my way correct ?
real-analysis calculus functions
real-analysis calculus functions
asked Dec 4 at 16:09
Testform
14
asked Dec 4 at 16:09
Testform
14
edited Dec 5 at 19:44
asked Dec 4 at 16:09
Testform
14
asked Dec 4 at 16:09
Testform
14
asked Dec 4 at 16:09
Testform
14
14
The question keeps changing and I cannot continue to update my answer. Please ask another question instead of changing it
– Federico
Dec 4 at 17:37
@Federico , Sorry, but this last shape, if I can take your comment, I will be glad.
– Testform
Dec 4 at 17:49
add a comment |
The question keeps changing and I cannot continue to update my answer. Please ask another question instead of changing it
– Federico
Dec 4 at 17:37
@Federico , Sorry, but this last shape, if I can take your comment, I will be glad.
– Testform
Dec 4 at 17:49
The question keeps changing and I cannot continue to update my answer. Please ask another question instead of changing it
– Federico
Dec 4 at 17:37
The question keeps changing and I cannot continue to update my answer. Please ask another question instead of changing it
– Federico
Dec 4 at 17:37
@Federico , Sorry, but this last shape, if I can take your comment, I will be glad.
– Testform
Dec 4 at 17:49
@Federico , Sorry, but this last shape, if I can take your comment, I will be glad.
– Testform
Dec 4 at 17:49
add a comment |
1 Answer
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Old version. Your first formula doesn't make much sense: the expression
$$
lim_{utoinfty} int_1^u g(x),dx
$$
is a constant, usually denoted as $int_1^infty g(x),dx$, not a function of $u$; therefore in this case its derivative would be $0$.
If instead you mean
$$
f(u) = int_1^u g(x),dx,
$$
then the Fundamental Theorem of Calculus tells you that
$$
f'(u) = g(u)
$$
for all $u$ at which $g$ is continuous; in your case $uinmathbb Rsetminusmathbb Z$.
At $uinmathbb Z$, your function $f$ admits left and right derivatives equal to the left and right limits $lim_{sto u^pm}g(s)$.
answered Dec 4 at 16:23
Federico
4,193512
Oh, that Leibniz rule! I thought you were referring to the other one.
– Federico
Dec 4 at 16:38
I added some details. You just need the FToC in this case, no need for differentiation under integral sign
– Federico
Dec 4 at 16:43
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
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oldest
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up vote
0
down vote
Old version. Your first formula doesn't make much sense: the expression
$$
lim_{utoinfty} int_1^u g(x),dx
$$
is a constant, usually denoted as $int_1^infty g(x),dx$, not a function of $u$; therefore in this case its derivative would be $0$.
If instead you mean
$$
f(u) = int_1^u g(x),dx,
$$
then the Fundamental Theorem of Calculus tells you that
$$
f'(u) = g(u)
$$
for all $u$ at which $g$ is continuous; in your case $uinmathbb Rsetminusmathbb Z$.
At $uinmathbb Z$, your function $f$ admits left and right derivatives equal to the left and right limits $lim_{sto u^pm}g(s)$.
answered Dec 4 at 16:23
Federico
4,193512
Oh, that Leibniz rule! I thought you were referring to the other one.
– Federico
Dec 4 at 16:38
I added some details. You just need the FToC in this case, no need for differentiation under integral sign
– Federico
Dec 4 at 16:43
add a comment |
up vote
0
down vote
Old version. Your first formula doesn't make much sense: the expression
$$
lim_{utoinfty} int_1^u g(x),dx
$$
is a constant, usually denoted as $int_1^infty g(x),dx$, not a function of $u$; therefore in this case its derivative would be $0$.
If instead you mean
$$
f(u) = int_1^u g(x),dx,
$$
then the Fundamental Theorem of Calculus tells you that
$$
f'(u) = g(u)
$$
for all $u$ at which $g$ is continuous; in your case $uinmathbb Rsetminusmathbb Z$.
At $uinmathbb Z$, your function $f$ admits left and right derivatives equal to the left and right limits $lim_{sto u^pm}g(s)$.
answered Dec 4 at 16:23
Federico
4,193512
Oh, that Leibniz rule! I thought you were referring to the other one.
– Federico
Dec 4 at 16:38
I added some details. You just need the FToC in this case, no need for differentiation under integral sign
– Federico
Dec 4 at 16:43
add a comment |
up vote
0
down vote
up vote
0
down vote
Old version. Your first formula doesn't make much sense: the expression
$$
lim_{utoinfty} int_1^u g(x),dx
$$
is a constant, usually denoted as $int_1^infty g(x),dx$, not a function of $u$; therefore in this case its derivative would be $0$.
If instead you mean
$$
f(u) = int_1^u g(x),dx,
$$
then the Fundamental Theorem of Calculus tells you that
$$
f'(u) = g(u)
$$
for all $u$ at which $g$ is continuous; in your case $uinmathbb Rsetminusmathbb Z$.
At $uinmathbb Z$, your function $f$ admits left and right derivatives equal to the left and right limits $lim_{sto u^pm}g(s)$.
answered Dec 4 at 16:23
Federico
4,193512
Old version. Your first formula doesn't make much sense: the expression
$$
lim_{utoinfty} int_1^u g(x),dx
$$
is a constant, usually denoted as $int_1^infty g(x),dx$, not a function of $u$; therefore in this case its derivative would be $0$.
If instead you mean
$$
f(u) = int_1^u g(x),dx,
$$
then the Fundamental Theorem of Calculus tells you that
$$
f'(u) = g(u)
$$
for all $u$ at which $g$ is continuous; in your case $uinmathbb Rsetminusmathbb Z$.
At $uinmathbb Z$, your function $f$ admits left and right derivatives equal to the left and right limits $lim_{sto u^pm}g(s)$.
answered Dec 4 at 16:23
Federico
4,193512
edited Dec 4 at 16:41
answered Dec 4 at 16:23
Federico
4,193512
answered Dec 4 at 16:23
Federico
4,193512
answered Dec 4 at 16:23
Federico
4,193512
4,193512
Oh, that Leibniz rule! I thought you were referring to the other one.
– Federico
Dec 4 at 16:38
I added some details. You just need the FToC in this case, no need for differentiation under integral sign
– Federico
Dec 4 at 16:43
add a comment |
Oh, that Leibniz rule! I thought you were referring to the other one.
– Federico
Dec 4 at 16:38
I added some details. You just need the FToC in this case, no need for differentiation under integral sign
– Federico
Dec 4 at 16:43
Oh, that Leibniz rule! I thought you were referring to the other one.
– Federico
Dec 4 at 16:38
Oh, that Leibniz rule! I thought you were referring to the other one.
– Federico
Dec 4 at 16:38
I added some details. You just need the FToC in this case, no need for differentiation under integral sign
– Federico
Dec 4 at 16:43
I added some details. You just need the FToC in this case, no need for differentiation under integral sign
– Federico
Dec 4 at 16:43
add a comment |
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The question keeps changing and I cannot continue to update my answer. Please ask another question instead of changing it
– Federico
Dec 4 at 17:37
@Federico , Sorry, but this last shape, if I can take your comment, I will be glad.
– Testform
Dec 4 at 17:49