derivative of integral (df/dt)^2*dt from 0 to x..

Question

amistre64 · Answer

the derivative of an integral gets you back to your innards as functions of x * f'

anonymous · Answer

what is the answer then..??

amistre64 · Answer

(dx/dt)^2 * x' has something to do with it, but im not sure what to make of the (df/dt)^2, id it meant as the square of the derivative or is it meant to be the 2nd derivative?

amistre64 · Answer

the answer is whatever it is, in the meantime .... what is the question?

anonymous · Answer

derivative of integral (df/dt)^2*dt from o to x i s asked

anonymous · Answer

it is square

amistre64 · Answer

hmm, well im not sure then how the ^2 effects the solution.  id have to review it and get back with you after my accounting exam

amistre64 · Answer

my gut says:

h(t) = (df/dt)^2

derivative of integrate h(t) 0 to x is h(x) x'  but then im just not confident yet how to redress it

anonymous · Answer

oki... however the answer is (df/dx)^2*x

amistre64 · Answer

if you know thats the correct answer, then i feel a bit more confident in my idea; change ts to xs and product rule.  not to sure how the x' becomes x tho :/

anonymous · Answer

(df/dx)^2 is the answer because of first integral theorem.