what's the derivative of the integral from x to 0 of x*f(t dt)?

Answer 1

Let

G(x) = Integral(0 to x, f(t) dt)

Then, by the Fundamental Theorem of Calculus,

G'(x) = f(x)

Answer 2

This is badly written. WITHOUT the "x*", i.e. if it's meant to be just the integral from x to 0 of (f(t) dt), the deivative would be - f(x), because the integral from x to 0 is MINUS the integral from 0 to x!

Therefore, it has to be the NEGATIVE of what you get by applying the Fundamental Theorem of Calculus, which assumes that x is the UPPER LIMIT, not the LOWER LIMIT of the integral.

In symbols, G(x) = Integral (x to 0) f(t) dt

= - Integral (0 to x) f(t) dt = - f(x). So:

G'(x) = - f(x).

Sorry, Puggy.

HOWEVER, if it's really the integral from x to 0 of x (f(t) dt), let's define THAT as H(x), i.e.:

H(x) = Integral(x to 0) of x (f(t) dt).

Then H'(x) = Integral(x to 0) of (f(t) dt) - x f(x),

or H'(x) = H(x) / x - x f(x),

where once again, the inverted order of the x-integration doesn't affect the first term on the RHS, written this way. However, it DOES affect the second term because THAT involves differentiating a limit of the integral, and it's "downstairs" instead of being "upstairs," as in the first example.

Live long and prosper.