help on the fundamental theorem of calculus, differentiating an integral?

I will use S for integral

The fundamental theorem says if f(y) = S from a to y (g(s)ds (where a is a constant) then df/dy=g(y) AA

First write your integral as S from 0 to x+ct - S from 0 to x-ct

Differentiating w.r.t x gives [g(x+ct) - g(x-ct)]/(2c), differentiating again w.r.tx gives [g'(x+ct)-g'(x-ct)]/2c

From AA, df/dt = (df/dy)(dy/dt) where y=x+ct in the first S and y=x-ct in the second

= [cg(x+ct)+cg(x-ct)]/(2c) = [g(x+ct)+g(x-ct)]/2

Differentiate again w.r.t. t , which you should now be able to do

2d elementary theorem f calculus says as long as f is non-stop on the oen era containing a then d/dx[integration from a to x of f(t) dt] =f(x) and elementary theorem of calculus states vital from a to b of f(x) dx = F(b) - F(a) ; for this reason the respond ought to be sqrt(a million + sec[x^24]) - sqrt(a million + sec[(-3x)^8] ....To the guy above me you're taking the by-product of the integrated function, the integrand IS the by-product of the antiderivative...occasion if i combine 2x dx i ger x^2 if i then take the by-product i'm getting returned 2x. d/dx[ integration from -3x to x^3 (a million+sec(t^8))dt] = F(x^3) -F(-3x) ; d/dx[F(x^3)] - d/dx[F(-3x)] = f(x^3) - f(-3x) ; you arent taking the by-product of a million+sec(x) you're taking the by-product of the mixture of it, that's in simple terms the integrand.