Re: calc question ,

In article <B3CUf.9423$8Y2.505@fed1read03>,
"quat" <spam@xxxxxxxx> wrote:

I am trying to show:

g(x,y,z) = I( g(xt, yt, zt), t, 0, 1) + I( [x*g_x(xt, yt, zt) + y*g_y(xt,
yt,zt) + z*g_z(xt, yt, zt)]t, t, 0, 1),

where I( f, x, 0, 1) means integral of f wrt x from 0 to 1. The book says
to differentiate t*g(xt, yt, zt) to see this. From the product rule I got:

(tg)' = g + tg'

But where to go from here? Also how to find g'? Is g(u) = g(u(x,y)) now?

It's false. Try it with g(x,y,z) = x. However, g(x,y,z) -
g(0,0,0) = I( [x*g_x(xt, yt, zt) + y*g_y(xt, yt,zt) + z*g_z(xt,
yt, zt)]t, t, 0, 1), simply by FTC.
.