Question

Differentiate:
v = 6t + 4 cos 2t.

Answer 1

piece of cake

Answer 2

\[6-8sin(2t)\]

Answer 3

v' = 6 -8cos(2t)

Answer 4

help me eating it/ ;)

Answer 5

6 - 8sin2t

Answer 6

@amistre: it will b sin*

Answer 7

did I not type a sin? lol

Answer 8

v' = 6 - 8 sin2t
since d/dx sin2t = -2cos2t

Answer 9

o whoops

Answer 10

i mean d/dx cos = -sin

Answer 11

my head said sin; and my fingers decide to go another route

Answer 12

can anybdy help me tht where can i find these formulaes?? :/

Answer 13

lol. @amisrte

Answer 14

the trig derives are as much memorization as the trig functions themselves

Answer 15

but; we can try to do it thru th elimit definition

Answer 16

where i can find its formulaes?

Answer 17

all of them.

Answer 18

\[\lim_{h->0}\frac{cos(x+h)-cos(x)}{x+h-x}\]

Answer 19

in the back or front of any calculus textbook id assume

Answer 20

\[cos(x)cos(h)-sin(x)sin(h)-cos(x)//h\]
\[cos(x)(cos(h)-1)-sin(x)sin(h)//h\]

Answer 21

\[\lim_{h->0} cos(x)\frac{cos(h-1)}{h}-\lim_{h->0}sin(x)\frac{sin(h)}{h}\]

Answer 22

\[{cos(h)\over h} ->0\]
\[\frac{sin(h)}{h}->1\]
leaning us with:
\[-sin(x)\]

Answer 23

k, thnks a TON! :D

Answer 24

had a typo up in there, but im sure you can see thru it ;)

Answer 25

rule of thumb; if it starts with a "c"; the derivative is "-"

cos = -sin
cot = -csc^2
csc = -csc cot

Answer 26

the others or just the evil twins .. so to speak

sin = cos
tan = sec^2
sec = sec tan

Answer 27

k, thnks a TON! :D

Answer 28

heyyy!

plz show me the working oif:
4 cos 2t

Answer 29

how u got:
8sin(2t)

Answer 30

well, it amounts to the 'chain rule' and the 'product rule' if you know the shortcuts.

(rl)' = r'l +rl'

and the chain rule is: F[g(x)] = F'[g(x)] * g'(x)

Answer 31

r=4 ; l=cos(2x) , F(g(x))=cos(2x) , g(x) = 2x
r'=0 ; l'= -sin(2x) * 2

[4 cos(2x)]' = 0(cos(2x)) +4(-2 sin(2x)) = -8sin(2x)

Answer 32

it looks more complicated than it is; some things just dont notate right on screen