Little extension of the Leibniz rule
$begingroup$
$ int_A^B F(A, x)dx $
Let's say I have an integral on this form. If F didn't depend on A (let's say it depends on C instead), the derivative of this integral with respect to A would simply be -F(C,A). However, since A is not only the boundary, but an argument on F, how can I take the derivative of this integral with respect to A?
calculus integration
asked Dec 21 '18 at 16:45
J. OzkJ. Ozk
11
$endgroup$
add a comment |
$begingroup$
$ int_A^B F(A, x)dx $
Let's say I have an integral on this form. If F didn't depend on A (let's say it depends on C instead), the derivative of this integral with respect to A would simply be -F(C,A). However, since A is not only the boundary, but an argument on F, how can I take the derivative of this integral with respect to A?
calculus integration
asked Dec 21 '18 at 16:45
J. OzkJ. Ozk
11
$endgroup$
add a comment |
$begingroup$
$ int_A^B F(A, x)dx $
Let's say I have an integral on this form. If F didn't depend on A (let's say it depends on C instead), the derivative of this integral with respect to A would simply be -F(C,A). However, since A is not only the boundary, but an argument on F, how can I take the derivative of this integral with respect to A?
calculus integration
asked Dec 21 '18 at 16:45
J. OzkJ. Ozk
11
$endgroup$
$ int_A^B F(A, x)dx $
Let's say I have an integral on this form. If F didn't depend on A (let's say it depends on C instead), the derivative of this integral with respect to A would simply be -F(C,A). However, since A is not only the boundary, but an argument on F, how can I take the derivative of this integral with respect to A?
calculus integration
calculus integration
asked Dec 21 '18 at 16:45
J. OzkJ. Ozk
11
asked Dec 21 '18 at 16:45
J. OzkJ. Ozk
11
edited Dec 21 '18 at 16:54
J. Ozk
asked Dec 21 '18 at 16:45
J. OzkJ. Ozk
11
asked Dec 21 '18 at 16:45
J. OzkJ. Ozk
11
asked Dec 21 '18 at 16:45
J. OzkJ. Ozk
11
11
add a comment |
add a comment |
1 Answer
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$begingroup$
Define $$f(s,t)=int_s^BF(t,x),dx.$$Say your integral is $I(A)$. Then $$I(A)=f(A,A),$$so the several-variable chain rule shows that $$I'(A)=frac{partial f}{partial s}(A,A)+frac{partial f}{partial t}(A,A).$$The first term is given by the Fundamental Theorem of Calculus, while the standard Leibniz rule gives the second term.
answered Dec 21 '18 at 17:01
David C. UllrichDavid C. Ullrich
61.6k43995
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Define $$f(s,t)=int_s^BF(t,x),dx.$$Say your integral is $I(A)$. Then $$I(A)=f(A,A),$$so the several-variable chain rule shows that $$I'(A)=frac{partial f}{partial s}(A,A)+frac{partial f}{partial t}(A,A).$$The first term is given by the Fundamental Theorem of Calculus, while the standard Leibniz rule gives the second term.
answered Dec 21 '18 at 17:01
David C. UllrichDavid C. Ullrich
61.6k43995
$endgroup$
add a comment |
$begingroup$
Define $$f(s,t)=int_s^BF(t,x),dx.$$Say your integral is $I(A)$. Then $$I(A)=f(A,A),$$so the several-variable chain rule shows that $$I'(A)=frac{partial f}{partial s}(A,A)+frac{partial f}{partial t}(A,A).$$The first term is given by the Fundamental Theorem of Calculus, while the standard Leibniz rule gives the second term.
answered Dec 21 '18 at 17:01
David C. UllrichDavid C. Ullrich
61.6k43995
$endgroup$
add a comment |
$begingroup$
Define $$f(s,t)=int_s^BF(t,x),dx.$$Say your integral is $I(A)$. Then $$I(A)=f(A,A),$$so the several-variable chain rule shows that $$I'(A)=frac{partial f}{partial s}(A,A)+frac{partial f}{partial t}(A,A).$$The first term is given by the Fundamental Theorem of Calculus, while the standard Leibniz rule gives the second term.
answered Dec 21 '18 at 17:01
David C. UllrichDavid C. Ullrich
61.6k43995
$endgroup$
Define $$f(s,t)=int_s^BF(t,x),dx.$$Say your integral is $I(A)$. Then $$I(A)=f(A,A),$$so the several-variable chain rule shows that $$I'(A)=frac{partial f}{partial s}(A,A)+frac{partial f}{partial t}(A,A).$$The first term is given by the Fundamental Theorem of Calculus, while the standard Leibniz rule gives the second term.
answered Dec 21 '18 at 17:01
David C. UllrichDavid C. Ullrich
61.6k43995
edited Dec 21 '18 at 17:24
answered Dec 21 '18 at 17:01
David C. UllrichDavid C. Ullrich
61.6k43995
answered Dec 21 '18 at 17:01
David C. UllrichDavid C. Ullrich
61.6k43995
answered Dec 21 '18 at 17:01
David C. UllrichDavid C. Ullrich
61.6k43995
61.6k43995
add a comment |
add a comment |
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