2007-08-24 07:30:54 · 7 answers · asked by Alexander 6 in Science & Mathematics ➔ Mathematics
x´(u)=1/u´(x) =-1/u^2(x)
d x´/du= dx´/dx* dx/du = -2u*u´/u^4 *(-1/u^2)=2/u^3 and at u=2
=1/4
2007-08-24 07:52:21 · answer #1 · answered by santmann2002 7 · 1⤊ 0⤋
I think you have du/dx + u^2 = 0, so that du/dx = -u^2, separating the variables you have du/(-u^2) = dx, and integrating gives
x(u) = 1/u
so that x'(u) = -1/u^2 and x''(u) = 2/u^3
plugging in u=2 gives 1/4, the answer above.
2007-08-24 14:54:08 · answer #2 · answered by Derek C 3 · 0⤊ 0⤋
1/4
2007-08-24 14:38:16 · answer #3 · answered by anotherhumanmale 5 · 1⤊ 1⤋
2(x)+4=0
2x + 4 = 0
2x=0 - 4
2x=-4
x=-4/2
x=-2
2007-08-24 14:48:14 · answer #4 · answered by smith cool 1 · 0⤊ 0⤋
57
2007-08-24 14:33:44 · answer #5 · answered by Anonymous · 1⤊ 1⤋
du/dx = -u^2
This means small change in u / small change in x
So dx/du = -1/u^2
d2x/du2 = 2/u^3
At u = 2, this value is 1/4
2007-08-24 16:06:08 · answer #6 · answered by Dr D 7 · 0⤊ 0⤋
Stop making me answer your math question you *****
2007-08-24 14:36:00 · answer #7 · answered by x3blackcats 2 · 0⤊ 1⤋