How to differentiate $ exp left{ int^t_sf(u)du right} $ w.r.t. $t$?
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I have looked and looked in my calculus books but couldn't find the answer to a problem I am facing of of the following form:
$$g(t) = exp left{ int^t_sf(u)du right} \
frac{partial g(t)}{partial t} = ? $$
How do I differentiate $g(t)$ w.r.t $t$?
$s$ is fixed.
calculus integration derivatives
edited Jan 14 at 10:43
BowPark
560720
asked Jan 13 at 21:15
econmajorreconmajorr
104
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add a comment |
$begingroup$
I have looked and looked in my calculus books but couldn't find the answer to a problem I am facing of of the following form:
$$g(t) = exp left{ int^t_sf(u)du right} \
frac{partial g(t)}{partial t} = ? $$
How do I differentiate $g(t)$ w.r.t $t$?
$s$ is fixed.
calculus integration derivatives
edited Jan 14 at 10:43
BowPark
560720
asked Jan 13 at 21:15
econmajorreconmajorr
104
$endgroup$
add a comment |
$begingroup$
I have looked and looked in my calculus books but couldn't find the answer to a problem I am facing of of the following form:
$$g(t) = exp left{ int^t_sf(u)du right} \
frac{partial g(t)}{partial t} = ? $$
How do I differentiate $g(t)$ w.r.t $t$?
$s$ is fixed.
calculus integration derivatives
edited Jan 14 at 10:43
BowPark
560720
asked Jan 13 at 21:15
econmajorreconmajorr
104
$endgroup$
I have looked and looked in my calculus books but couldn't find the answer to a problem I am facing of of the following form:
$$g(t) = exp left{ int^t_sf(u)du right} \
frac{partial g(t)}{partial t} = ? $$
How do I differentiate $g(t)$ w.r.t $t$?
$s$ is fixed.
calculus integration derivatives
calculus integration derivatives
edited Jan 14 at 10:43
BowPark
560720
asked Jan 13 at 21:15
econmajorreconmajorr
104
edited Jan 14 at 10:43
BowPark
560720
asked Jan 13 at 21:15
econmajorreconmajorr
104
edited Jan 14 at 10:43
BowPark
560720
edited Jan 14 at 10:43
BowPark
560720
edited Jan 14 at 10:43
BowPark
560720
560720
asked Jan 13 at 21:15
econmajorreconmajorr
104
asked Jan 13 at 21:15
econmajorreconmajorr
104
asked Jan 13 at 21:15
econmajorreconmajorr
104
104
add a comment |
add a comment |
3 Answers
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It is a composite of the function $h(t):= exp(t)$ and $k(t):= int_s^t f(u) , mathrm{d} u$. By the chain rule we have $g'(t) = k'(t) h'(k(t))$. On the other hand, the fundamental theorem of calculus says that $k'(t) = f(t)$. Both together imply that
$$g'(t) = k'(t) h'(k(t)) = f(t) exp(k(t)) = f(t) g(t).$$
Note that this argument also proves that $g$ is differentiable. This steps are valid provided that $f$ is contiuous.
answered Jan 13 at 21:19
p4schp4sch
5,450318
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add a comment |
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Chain rule; we know how to differentiate $exp$ and we know how to differentiate $int_s^t f(u),du$ (the Fundamental Theorem), so we just combine those with the chain rule..
answered Jan 13 at 21:19
jmerryjmerry
15.3k1632
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add a comment |
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We have $$frac{partial g(t)}{partial t}{=expleft{ int^t_sf(u)du right}cdot {partial int^t_sf(u)duover partial t}\=g(t)cdot Big(f(t)-f(s)Big)}$$can't proceed any further.
answered Jan 13 at 22:47
Mostafa AyazMostafa Ayaz
17k31039
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add a comment |
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3 Answers
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active
oldest
votes
3 Answers
3
active
oldest
votes
active
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active
oldest
votes
$begingroup$
It is a composite of the function $h(t):= exp(t)$ and $k(t):= int_s^t f(u) , mathrm{d} u$. By the chain rule we have $g'(t) = k'(t) h'(k(t))$. On the other hand, the fundamental theorem of calculus says that $k'(t) = f(t)$. Both together imply that
$$g'(t) = k'(t) h'(k(t)) = f(t) exp(k(t)) = f(t) g(t).$$
Note that this argument also proves that $g$ is differentiable. This steps are valid provided that $f$ is contiuous.
answered Jan 13 at 21:19
p4schp4sch
5,450318
$endgroup$
add a comment |
$begingroup$
It is a composite of the function $h(t):= exp(t)$ and $k(t):= int_s^t f(u) , mathrm{d} u$. By the chain rule we have $g'(t) = k'(t) h'(k(t))$. On the other hand, the fundamental theorem of calculus says that $k'(t) = f(t)$. Both together imply that
$$g'(t) = k'(t) h'(k(t)) = f(t) exp(k(t)) = f(t) g(t).$$
Note that this argument also proves that $g$ is differentiable. This steps are valid provided that $f$ is contiuous.
answered Jan 13 at 21:19
p4schp4sch
5,450318
$endgroup$
add a comment |
$begingroup$
It is a composite of the function $h(t):= exp(t)$ and $k(t):= int_s^t f(u) , mathrm{d} u$. By the chain rule we have $g'(t) = k'(t) h'(k(t))$. On the other hand, the fundamental theorem of calculus says that $k'(t) = f(t)$. Both together imply that
$$g'(t) = k'(t) h'(k(t)) = f(t) exp(k(t)) = f(t) g(t).$$
Note that this argument also proves that $g$ is differentiable. This steps are valid provided that $f$ is contiuous.
answered Jan 13 at 21:19
p4schp4sch
5,450318
$endgroup$
It is a composite of the function $h(t):= exp(t)$ and $k(t):= int_s^t f(u) , mathrm{d} u$. By the chain rule we have $g'(t) = k'(t) h'(k(t))$. On the other hand, the fundamental theorem of calculus says that $k'(t) = f(t)$. Both together imply that
$$g'(t) = k'(t) h'(k(t)) = f(t) exp(k(t)) = f(t) g(t).$$
Note that this argument also proves that $g$ is differentiable. This steps are valid provided that $f$ is contiuous.
answered Jan 13 at 21:19
p4schp4sch
5,450318
edited Jan 14 at 10:10
answered Jan 13 at 21:19
p4schp4sch
5,450318
answered Jan 13 at 21:19
p4schp4sch
5,450318
answered Jan 13 at 21:19
p4schp4sch
5,450318
5,450318
add a comment |
add a comment |
$begingroup$
Chain rule; we know how to differentiate $exp$ and we know how to differentiate $int_s^t f(u),du$ (the Fundamental Theorem), so we just combine those with the chain rule..
answered Jan 13 at 21:19
jmerryjmerry
15.3k1632
$endgroup$
add a comment |
$begingroup$
Chain rule; we know how to differentiate $exp$ and we know how to differentiate $int_s^t f(u),du$ (the Fundamental Theorem), so we just combine those with the chain rule..
answered Jan 13 at 21:19
jmerryjmerry
15.3k1632
$endgroup$
add a comment |
$begingroup$
Chain rule; we know how to differentiate $exp$ and we know how to differentiate $int_s^t f(u),du$ (the Fundamental Theorem), so we just combine those with the chain rule..
answered Jan 13 at 21:19
jmerryjmerry
15.3k1632
$endgroup$
Chain rule; we know how to differentiate $exp$ and we know how to differentiate $int_s^t f(u),du$ (the Fundamental Theorem), so we just combine those with the chain rule..
answered Jan 13 at 21:19
jmerryjmerry
15.3k1632
answered Jan 13 at 21:19
jmerryjmerry
15.3k1632
answered Jan 13 at 21:19
jmerryjmerry
15.3k1632
answered Jan 13 at 21:19
jmerryjmerry
15.3k1632
15.3k1632
add a comment |
add a comment |
$begingroup$
We have $$frac{partial g(t)}{partial t}{=expleft{ int^t_sf(u)du right}cdot {partial int^t_sf(u)duover partial t}\=g(t)cdot Big(f(t)-f(s)Big)}$$can't proceed any further.
answered Jan 13 at 22:47
Mostafa AyazMostafa Ayaz
17k31039
$endgroup$
add a comment |
$begingroup$
We have $$frac{partial g(t)}{partial t}{=expleft{ int^t_sf(u)du right}cdot {partial int^t_sf(u)duover partial t}\=g(t)cdot Big(f(t)-f(s)Big)}$$can't proceed any further.
answered Jan 13 at 22:47
Mostafa AyazMostafa Ayaz
17k31039
$endgroup$
add a comment |
$begingroup$
We have $$frac{partial g(t)}{partial t}{=expleft{ int^t_sf(u)du right}cdot {partial int^t_sf(u)duover partial t}\=g(t)cdot Big(f(t)-f(s)Big)}$$can't proceed any further.
answered Jan 13 at 22:47
Mostafa AyazMostafa Ayaz
17k31039
$endgroup$
We have $$frac{partial g(t)}{partial t}{=expleft{ int^t_sf(u)du right}cdot {partial int^t_sf(u)duover partial t}\=g(t)cdot Big(f(t)-f(s)Big)}$$can't proceed any further.
answered Jan 13 at 22:47
Mostafa AyazMostafa Ayaz
17k31039
answered Jan 13 at 22:47
Mostafa AyazMostafa Ayaz
17k31039
answered Jan 13 at 22:47
Mostafa AyazMostafa Ayaz
17k31039
answered Jan 13 at 22:47
Mostafa AyazMostafa Ayaz
17k31039
17k31039
add a comment |
add a comment |
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