What is the derivative of an integral of xf(t)dt from 0 to x. HINT: the answer is not xf(x).?

Answer 1

Let

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

That means we can pull the x out of the integral, since the integral is expressed in terms of dt.

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

To take the derivative, we use the product rule.

G'(x) = Integral(0 to x, f(t) dt) + x(f(x))

If you want it expressed as a differentiation equation, it's equal to

G'(x) = G(x) + x f(x)

Answer 2

I[0,x](xf(t)dt) = x*I[0,x]( f(t)dt )
derivative =d/dx(x*I[0,x]( f(t)dt ))
= xf(x) + I[0,x]( f(t)dt ) .....chain rule

and Dr. Spock Blabbers too much on a wrong ans.

Answer 3

we can coach by using induction that the nth by-made of e^(-a million/x^2) has the variety g_n(x) * e^(-a million/x^2) the place g_n(x) is a rational function. Then we use the shrink definition of by-product, and induction on n. If f^(n)(x) exists for all x, then f^(n+a million)(0) = lim x->0 f^(n)(x)/x = lim x->0 g_n(x)*e^(-a million/x^2)/x. This shrink is 0 because of the fact e^(-a million/x^2) methods 0 quicker than the denominator of g_n(x). EDIT: Oops, John D published a miles better answer mutually as i exchange into composing mine.

Answer 4

let
y = x(F(x) - F(0)) = ∫xf(t)dt from 0 to x
then
dy/dx = (F(x) - F(0)) + x(F'(x) - F'(0))

Answer 5

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

G'(x) = Integral(0 to x, f(t))dt + x f(x) --- or, in Puggy's terms,

G'(x) = G(x) / x + x f(x). [His corresponding first term on the RHS lacks the needed "/ x" divisor.]

Live long and prosper.

Answer 6

it's xf(t)