How to Find the Derivative of an Integral Using Differentiation

[Niles]

Homework Statement

Hi

Say I have for example

$$f(x) = \int_0^x {e^{ - (x - x')} g(x')\,dx'}$$

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

$$f'(x) = g'(x) - g(x)$$

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

Best.

 

[HallsofIvy]

No, it is not.

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

In your problem, $\beta(x)= x$, $\alpha(x)= 0$, and $f(x,x')= e^{x-x'}g(x')$.

 

[Niles]

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 fundamental theorem of calculus?

 

[eczeno]

hello,

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

$$f(x) = \int_0^x{g(t)}dt$$

then $$f'(x) = g(x)$$ .

but your function is of the form:

$$f(x) = \int_0^x{h(x,t) g(t)}dt$$

which is more complicated.

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

hope this helps.

 

[Niles]

I see, thanks to both of you.

Best.

 

[eczeno]

cheers niles,

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