Question

Use the Product and Quotient Rules as necessary to find the derivative of the given function.

f (x) = 3x^3 cos(x)

Answer 1

This is what I did, but I got it wrong.
\[3x^3\cos(x)\]
\[9x^2-\sin(x)\]

Answer 2

Factor out constants:
= 3 (d/dx(x^3 cos(x)))
Use the product rule, d/dx(u v) = v ( du)/( dx)+u ( dv)/( dx), where u = x^3 and v = cos(x):
= 3 (x^3 (d/dx(cos(x)))+cos(x) (d/dx(x^3)))
The derivative of x^3 is 3 x^2:
= 3 (x^3 (d/dx(cos(x)))+(3 x^2) cos(x))
The derivative of cos(x) is -sin(x):
= 3 (x^3 -sin(x)+3 x^2 cos(x))

Answer 3

when u take the derivative of one function, th esecond function remains as it is

Answer 4

9x^2cos(x)+3x^3(-sinx)

Answer 5

Corey89, that was just a big blob of idk. Can you clean that up?

Answer 6

uzma, can you show the formula you used to do that?

Answer 7

(f.g)'=f'g+fg'

Answer 8

I'm really lost.

Answer 9

where prime mean derivative

Answer 10

factor out the constants, which in this case is 3, so f(x) = 3(x^3cos(x))

Answer 11

\[(cfg)'=c(fg)'=c(f'g+fg')\]

Answer 12

How did you get what f was and what g was?

Answer 13

c=3
f=x^3
g=cos(x)

Answer 14

C?!?!?!

Answer 15

the product rule is (f*g) = f'*g + f*g' where ' is the derivative

Answer 16

or
you could say
c=3
f=cos(x)
g=x^3

Answer 17

f would be 3x^3 and g would be cos x

Answer 18

c is a constant

Answer 19

I can't follow with 3 people. Can ONE person just do this step by step with nice spaces and neat and whatnot?

Answer 20

:)...ok myiniaya u go

Answer 21

I don't see why I need to factor out 3?

Answer 22

you don't need to

Answer 23

hammafer is really lost

Answer 24

m so bad in typing :)

Answer 25

Ok, so I just need to do f'g+g'f, correct?

Answer 26

yes

Answer 27

\[y=3x^3\cos(x)=cf(x)g(x) => c=3, f(x)=x^3=>g(x)=\cos(x)\]
\[y'=c(f'g+fg')\]

Answer 28

Do I need a constant?

Answer 29

3 is a constant multiple

Answer 30

you can do f=3x^3 and g=cos(x) if you want

Answer 31

you can do f=x^3 and g=3cos(x) if you want

Answer 32

I prefer doing it the second way, myininaya

Answer 33

what 2nd way

Answer 34

Wait.

Answer 35

you can do f=3x^3 and g=cos(x) if you want

Answer 36

That one.

Answer 37

then f'=9x^2 and g'=-sin(x)
plug in to the following formula:
f'g+fg'

Answer 38

SO
\[(9x^2)(\cos(x))+(3x^3)(-\sin(x))\]

Answer 39

yes

Answer 40

Then what?

Answer 41

this is the derivative

Answer 42

That's the answer?

Answer 43

yes you can rearrange some stuff but it is the answer

Answer 44

How do I rearrange it?

Answer 45

\[3x^2(3\cos(x)-x \sin(x))\]

Answer 46

may i say something pls?

Answer 47

hamm...y dont u post ur work to be checked for error, otherwise u'll be lost more