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







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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















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







share|cite|improve this question
























  • 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













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







share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













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share|cite|improve this question








edited Dec 5 at 19:44

























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


















  • 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










1 Answer
1






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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 $lim_{sto u^pm}g(s)$.






share|cite|improve this answer























  • 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











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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 $lim_{sto u^pm}g(s)$.






share|cite|improve this answer























  • 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















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 $lim_{sto u^pm}g(s)$.






share|cite|improve this answer























  • 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













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 $lim_{sto u^pm}g(s)$.






share|cite|improve this answer















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 $lim_{sto u^pm}g(s)$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 4 at 16:41

























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


















  • 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


















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