2007-09-21 05:36:13 · 11 answers · asked by nala 1 in Science & Mathematics ➔ Mathematics
d(t^3) = 3t^2
d(2t) = 2
d(4) = 0
d(t^3 - 2t + 4) = 3t^2 - 2 + 0
= 3t^2 - 2
2007-09-21 05:43:20 · answer #1 · answered by mohanrao d 7 · 0⤊ 0⤋
To find the derivative, take the exponent, mutiply that by the coefficient; then subtract one from the exponent. Do this for each number.
So t^3 = 3t^2 (or (3*1) t ^ (3-1))
-2t can also be written as -2t^1, so the derivative is -2t^0 or -2 (because t^0 is also 1)
The four doesn't have an exponent, it is a constant, so it is 0
Ans: 3t^2-2
2007-09-21 14:33:36 · answer #2 · answered by Tina R 4 · 0⤊ 0⤋
To do derivative - you take the coefficents (the little numbers or the ^ numbers) and place them in the front of a term. You then subtract 1. So t^4 first der = 4t^3.
Lets look at your problem
first der of t^3-2t+4 = 3t^2 - 2t^0 + 0 note (no coefficent item goes away)
answer 3t^2 - 2 (note t^0 =1)
make sense? Hope thjis helps.
2007-09-21 12:44:16 · answer #3 · answered by pyz01 7 · 0⤊ 0⤋
s(t)=t^3-2t+4
s`(t)=3t^2-2
2007-09-21 12:44:11 · answer #4 · answered by Anonymous · 0⤊ 0⤋
s(t)=t^3-2t+4
s'(t)=3t^2-2
2007-09-21 12:45:11 · answer #5 · answered by Cranky 2 · 0⤊ 0⤋
3t^2 -2
2007-09-21 12:41:02 · answer #6 · answered by Anonymous · 0⤊ 0⤋
3t^2-2
2007-09-21 12:56:02 · answer #7 · answered by savager 1 · 0⤊ 0⤋
3t^2-2
2007-09-21 12:45:06 · answer #8 · answered by ironduke8159 7 · 0⤊ 0⤋
3t^2-2
2007-09-21 12:41:04 · answer #9 · answered by vhentesiete 1 · 0⤊ 0⤋
s'(t) = 3t² - 2
2007-09-21 12:40:36 · answer #10 · answered by Dave 6 · 0⤊ 0⤋