Differentiate an integral

  • Thread starter Niles
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  • #1
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Homework Statement


Hi

Say I have for example

[tex]
f(x) = \int_0^x {e^{ - (x - x')} g(x')\,dx'}
[/tex]

Then is it correct that the derivative of f(x), f'(x), is given by

[tex]
f'(x) = g'(x) - g(x)
[/tex]

obtained by differentiating the integrand, and evaluate the result at x'=x?

Best.
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
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No, it is not.

Laplace's formula:
[tex]\frac{d}{dx}\int_{\alpha(x)}^{\beta(x)} f(x,y)dy= f(x, \beta(x))\frac{d\beta}{dx}- f(x,\alpha(x))\frac{d\alpha}{dx}+ \int_{\alpha(x)}^{\beta(x)} \frac{\partial f}{\partial x} dy[/tex]

In your problem, [itex]\beta(x)= x[/itex], [itex]\alpha(x)= 0[/itex], and [itex]f(x,x')= e^{x-x'}g(x')[/itex].
 
  • #3
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Pretty impressive that you managed to read and reply in ~3 minutes. A tip of the hat to you!

Is that the most general way? Or does it all come down to the fundemental theorem of calculus?
 
  • #4
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hello,

that is not quite correct. the fundamental theorem of calculus tells us that if

[tex] f(x) = \int_0^x{g(t)}dt[/tex]

then [tex]f'(x) = g(x)[/tex] .

but your function is of the form:

[tex] f(x) = \int_0^x{h(x,t) g(t)}dt[/tex]

which is more complicated.

try integration by parts on the right hand side before differentiating.

hope this helps.
 
  • #5
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I see, thanks to both of you.

Best.
 
  • #6
242
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cheers niles,

p.s. i believe if you try my method you will see the solution falls out very nicely and quickly.
 

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