Compute the derivate of Q(x)=2x^3, using the limit definition. just before you subsititue h=0, what expression are you taking the limit of/
2006-10-23 10:10:59 · 6 answers · asked by MJ 1 in Science & Mathematics ➔ Mathematics
Q'(x)=6x^2
Q(x)=2x^3
y=2x^3
y+delta y=2(x+h)^3
delta y=2[(x+h)^3-x^3]
dy/dx=lim h>0 2[(x+h)^3-x^3]/h
=lim h>0 2[x^3+3x^2h+3xh^2+h^3-x^3]/h
=limh>0 2[3x^2h/h+3xh^2/h+h^3/h]
allowing h=0
=2[3x^2]=6x^2
2006-10-23 10:19:00 · answer #1 · answered by raj 7 · 0⤊ 0⤋
the derivative is 6x^2
the exponent 3 gets multiplied times the 2 and the exponent gets 1 subtracted from it
2006-10-23 17:16:34 · answer #2 · answered by devilishblueyes 7 · 0⤊ 0⤋
the Derivitive of 2x^3 is 6x^2
2006-10-23 18:01:59 · answer #3 · answered by vabanu 2 · 0⤊ 0⤋
Q(x+h) = 2(x+h)^3
Q(x+h) = 2x3 + 6x2h + 6xh2 + 2h3
Q(x) = 2x3
Q(x+h) - Q(x) = 6x2h + 6xh2 + 2h3
[Q(x+h) - Q(x)]/h = 6x2 + 6xh + 2h2
If h-->0 then [Q(x+h) - Q(x)]/h --> 6x2
Th
2006-10-23 17:35:56 · answer #4 · answered by Thermo 6 · 0⤊ 0⤋
blue?
2006-10-23 17:14:04 · answer #5 · answered by Anonymous · 0⤊ 0⤋
you find it
2006-10-23 17:12:40 · answer #6 · answered by Anonymous · 0⤊ 0⤋