2007-11-12 14:51:48 · 5 answers · asked by aminer_lover90 1 in Science & Mathematics ➔ Mathematics
Piece of cake, Ms. aminer_l. Let t= (3-3u) and dt/du= -3. Then g(t) = 2 * t^(-1/2)
We use power rule and chain rule together to find 1st and 2d derivitives:
g'(u) = d g(t)/dt x dt/du = 2 x (-1/2) x (-3) t^(-3/2)
= -3 t^(-3/2)
Do the same for the second derivative.
2007-11-12 15:03:20 · answer #1 · answered by cattbarf 7 · 0⤊ 0⤋
g=2(3-3u)^(-1/2)
g' = 2(-1/2)(3-3u)^(-3/2)(-3) = (3)(3-3u)^(-3/2)
g'' = (3)(-3/2)(3-3u)^(-5/2)(-3) = (27/2)(3-3u)^(-5/2)
2007-11-12 15:00:58 · answer #2 · answered by dooner75 3 · 0⤊ 0⤋
g'(u) = 1/[sqrt(3-3u)]
g''(u) = -1/[2*sqrt([3-3u]^3)]
2007-11-12 14:57:21 · answer #3 · answered by Zmik 3 · 0⤊ 1⤋
g(u)=2/(3-3u)^1/2
2*(3-3u)^-1/2
g(u)'=-(3)^-3/2
g(u)"=1
2007-11-12 14:59:57 · answer #4 · answered by Faisal R 3 · 0⤊ 1⤋
seven.
2007-11-12 14:53:32 · answer #5 · answered by Anonymous · 0⤊ 2⤋