What ins the derivative of int_0^tvarphi(t,u)du
Let $f(t)=int_0^tvarphi(t,u)du$ where $varphi$ is a continuos function. Do yo know how to compute the derivative of f?
It is not possible to directly apply the fundamental theorem of calculus.
multivariable-calculus
asked Nov 30 at 14:15
Adrián Hinojosa Calleja
937
add a comment |
Let $f(t)=int_0^tvarphi(t,u)du$ where $varphi$ is a continuos function. Do yo know how to compute the derivative of f?
It is not possible to directly apply the fundamental theorem of calculus.
multivariable-calculus
asked Nov 30 at 14:15
Adrián Hinojosa Calleja
937
1
Do you mean the derivative of $f$?
– Olivier Moschetta
Nov 30 at 14:17
add a comment |
Let $f(t)=int_0^tvarphi(t,u)du$ where $varphi$ is a continuos function. Do yo know how to compute the derivative of f?
It is not possible to directly apply the fundamental theorem of calculus.
multivariable-calculus
asked Nov 30 at 14:15
Adrián Hinojosa Calleja
937
Let $f(t)=int_0^tvarphi(t,u)du$ where $varphi$ is a continuos function. Do yo know how to compute the derivative of f?
It is not possible to directly apply the fundamental theorem of calculus.
multivariable-calculus
multivariable-calculus
asked Nov 30 at 14:15
Adrián Hinojosa Calleja
937
asked Nov 30 at 14:15
Adrián Hinojosa Calleja
937
edited Nov 30 at 18:11
asked Nov 30 at 14:15
Adrián Hinojosa Calleja
937
asked Nov 30 at 14:15
Adrián Hinojosa Calleja
937
asked Nov 30 at 14:15
Adrián Hinojosa Calleja
937
937
1
Do you mean the derivative of $f$?
– Olivier Moschetta
Nov 30 at 14:17
add a comment |
1
Do you mean the derivative of $f$?
– Olivier Moschetta
Nov 30 at 14:17
1
1
Do you mean the derivative of $f$?
– Olivier Moschetta
Nov 30 at 14:17
Do you mean the derivative of $f$?
– Olivier Moschetta
Nov 30 at 14:17
add a comment |
1 Answer
1
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I'm assuming you meant the derivative of $f$ in your post.
In a simple setting we need some assumptions on $varphi$ namely that $dfrac{partialvarphi}{partial t}$ exists and is continuous (to simplify things). Then the derivative of $f$ is obtained by
$$f'(t)=int_0^t dfrac{partialvarphi}{partial t}(t,u),du+varphi(t,t)$$
We get two bits, in the first one we differentiate under the integral sign while the second looks like an application of the fundamental theorem of calculus.
Note that the integral above is well-defined since $dfrac{partialvarphi}{partial t}$ is continuous.
answered Nov 30 at 14:21
Olivier Moschetta
2,7761411
1
You mean $varphi(t,t)$ and $varphi(t,0)$. But the last one should not appear in my opinion.
– Dog_69
Nov 30 at 14:24
Thanks! I'll correct.
– Olivier Moschetta
Nov 30 at 14:25
1
I don't understand why you write $varphi(t,0)$. In a single variable the theorem States that $d/dt int_0^t f(u)du= f(t)$. The value of $f$ at the initial point doesn't matter.
– Dog_69
Nov 30 at 19:26
You're right, it shouldn't be there. I wrongly applied a more general result where the bounds of the integral are two functions $g(t)$ and $h(t)$, then you get one term for each bound.
– Olivier Moschetta
Dec 1 at 15:42
Okey. Perfect. :)
– Dog_69
Dec 1 at 21:02
add a comment |
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I'm assuming you meant the derivative of $f$ in your post.
In a simple setting we need some assumptions on $varphi$ namely that $dfrac{partialvarphi}{partial t}$ exists and is continuous (to simplify things). Then the derivative of $f$ is obtained by
$$f'(t)=int_0^t dfrac{partialvarphi}{partial t}(t,u),du+varphi(t,t)$$
We get two bits, in the first one we differentiate under the integral sign while the second looks like an application of the fundamental theorem of calculus.
Note that the integral above is well-defined since $dfrac{partialvarphi}{partial t}$ is continuous.
answered Nov 30 at 14:21
Olivier Moschetta
2,7761411
1
You mean $varphi(t,t)$ and $varphi(t,0)$. But the last one should not appear in my opinion.
– Dog_69
Nov 30 at 14:24
Thanks! I'll correct.
– Olivier Moschetta
Nov 30 at 14:25
1
I don't understand why you write $varphi(t,0)$. In a single variable the theorem States that $d/dt int_0^t f(u)du= f(t)$. The value of $f$ at the initial point doesn't matter.
– Dog_69
Nov 30 at 19:26
You're right, it shouldn't be there. I wrongly applied a more general result where the bounds of the integral are two functions $g(t)$ and $h(t)$, then you get one term for each bound.
– Olivier Moschetta
Dec 1 at 15:42
Okey. Perfect. :)
– Dog_69
Dec 1 at 21:02
add a comment |
I'm assuming you meant the derivative of $f$ in your post.
In a simple setting we need some assumptions on $varphi$ namely that $dfrac{partialvarphi}{partial t}$ exists and is continuous (to simplify things). Then the derivative of $f$ is obtained by
$$f'(t)=int_0^t dfrac{partialvarphi}{partial t}(t,u),du+varphi(t,t)$$
We get two bits, in the first one we differentiate under the integral sign while the second looks like an application of the fundamental theorem of calculus.
Note that the integral above is well-defined since $dfrac{partialvarphi}{partial t}$ is continuous.
answered Nov 30 at 14:21
Olivier Moschetta
2,7761411
1
You mean $varphi(t,t)$ and $varphi(t,0)$. But the last one should not appear in my opinion.
– Dog_69
Nov 30 at 14:24
Thanks! I'll correct.
– Olivier Moschetta
Nov 30 at 14:25
1
I don't understand why you write $varphi(t,0)$. In a single variable the theorem States that $d/dt int_0^t f(u)du= f(t)$. The value of $f$ at the initial point doesn't matter.
– Dog_69
Nov 30 at 19:26
You're right, it shouldn't be there. I wrongly applied a more general result where the bounds of the integral are two functions $g(t)$ and $h(t)$, then you get one term for each bound.
– Olivier Moschetta
Dec 1 at 15:42
Okey. Perfect. :)
– Dog_69
Dec 1 at 21:02
add a comment |
I'm assuming you meant the derivative of $f$ in your post.
In a simple setting we need some assumptions on $varphi$ namely that $dfrac{partialvarphi}{partial t}$ exists and is continuous (to simplify things). Then the derivative of $f$ is obtained by
$$f'(t)=int_0^t dfrac{partialvarphi}{partial t}(t,u),du+varphi(t,t)$$
We get two bits, in the first one we differentiate under the integral sign while the second looks like an application of the fundamental theorem of calculus.
Note that the integral above is well-defined since $dfrac{partialvarphi}{partial t}$ is continuous.
answered Nov 30 at 14:21
Olivier Moschetta
2,7761411
I'm assuming you meant the derivative of $f$ in your post.
In a simple setting we need some assumptions on $varphi$ namely that $dfrac{partialvarphi}{partial t}$ exists and is continuous (to simplify things). Then the derivative of $f$ is obtained by
$$f'(t)=int_0^t dfrac{partialvarphi}{partial t}(t,u),du+varphi(t,t)$$
We get two bits, in the first one we differentiate under the integral sign while the second looks like an application of the fundamental theorem of calculus.
Note that the integral above is well-defined since $dfrac{partialvarphi}{partial t}$ is continuous.
answered Nov 30 at 14:21
Olivier Moschetta
2,7761411
edited Dec 1 at 15:42
answered Nov 30 at 14:21
Olivier Moschetta
2,7761411
answered Nov 30 at 14:21
Olivier Moschetta
2,7761411
answered Nov 30 at 14:21
Olivier Moschetta
2,7761411
2,7761411
1
You mean $varphi(t,t)$ and $varphi(t,0)$. But the last one should not appear in my opinion.
– Dog_69
Nov 30 at 14:24
Thanks! I'll correct.
– Olivier Moschetta
Nov 30 at 14:25
1
I don't understand why you write $varphi(t,0)$. In a single variable the theorem States that $d/dt int_0^t f(u)du= f(t)$. The value of $f$ at the initial point doesn't matter.
– Dog_69
Nov 30 at 19:26
You're right, it shouldn't be there. I wrongly applied a more general result where the bounds of the integral are two functions $g(t)$ and $h(t)$, then you get one term for each bound.
– Olivier Moschetta
Dec 1 at 15:42
Okey. Perfect. :)
– Dog_69
Dec 1 at 21:02
add a comment |
1
You mean $varphi(t,t)$ and $varphi(t,0)$. But the last one should not appear in my opinion.
– Dog_69
Nov 30 at 14:24
Thanks! I'll correct.
– Olivier Moschetta
Nov 30 at 14:25
1
I don't understand why you write $varphi(t,0)$. In a single variable the theorem States that $d/dt int_0^t f(u)du= f(t)$. The value of $f$ at the initial point doesn't matter.
– Dog_69
Nov 30 at 19:26
You're right, it shouldn't be there. I wrongly applied a more general result where the bounds of the integral are two functions $g(t)$ and $h(t)$, then you get one term for each bound.
– Olivier Moschetta
Dec 1 at 15:42
Okey. Perfect. :)
– Dog_69
Dec 1 at 21:02
1
1
You mean $varphi(t,t)$ and $varphi(t,0)$. But the last one should not appear in my opinion.
– Dog_69
Nov 30 at 14:24
You mean $varphi(t,t)$ and $varphi(t,0)$. But the last one should not appear in my opinion.
– Dog_69
Nov 30 at 14:24
Thanks! I'll correct.
– Olivier Moschetta
Nov 30 at 14:25
Thanks! I'll correct.
– Olivier Moschetta
Nov 30 at 14:25
1
1
I don't understand why you write $varphi(t,0)$. In a single variable the theorem States that $d/dt int_0^t f(u)du= f(t)$. The value of $f$ at the initial point doesn't matter.
– Dog_69
Nov 30 at 19:26
I don't understand why you write $varphi(t,0)$. In a single variable the theorem States that $d/dt int_0^t f(u)du= f(t)$. The value of $f$ at the initial point doesn't matter.
– Dog_69
Nov 30 at 19:26
You're right, it shouldn't be there. I wrongly applied a more general result where the bounds of the integral are two functions $g(t)$ and $h(t)$, then you get one term for each bound.
– Olivier Moschetta
Dec 1 at 15:42
You're right, it shouldn't be there. I wrongly applied a more general result where the bounds of the integral are two functions $g(t)$ and $h(t)$, then you get one term for each bound.
– Olivier Moschetta
Dec 1 at 15:42
Okey. Perfect. :)
– Dog_69
Dec 1 at 21:02
Okey. Perfect. :)
– Dog_69
Dec 1 at 21:02
add a comment |
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1
Do you mean the derivative of $f$?
– Olivier Moschetta
Nov 30 at 14:17