Question

What is the nth derivative of sin^3x?

Answer 1

what is the first derivative of your expression?

Answer 2

3/4(cosx- cos3x)

Answer 3

http://en.wikipedia.org/wiki/Fa%C3%A0_di_Bruno's_formula

Answer 4

i want the answer

Answer 5

is your problem \(\large \sin^3 (x) \)?

Answer 6

ya

Answer 7

Use the chain rule. \[\large \sin^3 (x) = [\sin(x)]^3\]\[\large \frac{d}{dx}[\sin (x)]^3 * \frac{d}{dx}[\sin(x)]\]

Answer 8

Why is this question still open??!

Answer 9

i want the full solution.

Answer 10

LOL

Answer 11

you mean you havent tried solving it for the past week?!!? cmon...

Answer 12

Using Bruno's Formula isn't easy. It's a lot of work. But one would use that if you were to actually provide a solution unless there is an easier way.

Answer 13

Maybe diff. this is better?!\[sin(3x) =3sinx - 4sin^3(x)\]\[sin^3(x) = \frac{3sin(x)-sin(3x)}{4}\]

Answer 14

@Callisto what's that for?

Answer 15

Since \(sin^3(x) = \frac{3sin(x)-sin(3x)}{4}\)
Differentiate sin^3(x) is the same as differentiate \(\frac{3sin(x)-sin(3x)}{4}\)
To save some work . If you keep differentiate sin^3x, not only need you apply chain rule, you have to apply product rules in finding derivatives later. If you differentiate \(\frac{3sin(x)-sin(3x)}{4}\), all you need is chain rule (for sin(3x)), it becomes easier.

Answer 16

how's that easier lol. you will be going through sin and cosine as you keep differentiating forever regardless of what you differentiate. In both situations you have to use the chain rule.

Answer 17

But you don't need product rule :D

Answer 18

\[\frac{d}{dx}sin^3x\]\[=3sin^2xcosx\]Then you need product rule when you diff. it again

Answer 19

i want the nth derivative.can somebody help me wid the solution?

Answer 20

@Sidhulucky Diff. it and observe the pattern.

Answer 21

@Callisto there is no pattern.

Answer 22

other than that you will keep hopping between sine and cosine as you take higher derivatives.

Answer 23

That's the pattern

Answer 24

http://answers.yahoo.com/question/index?qid=20091013111456AA7WCxa

Answer 25

lol what you show there is basic sine and cosine without any powers. same doesn't go for this case.

Answer 26

That's why we need \[sin^3(x) = \frac{3sin(x)-sin(3x)}{4}\]

Answer 27

If we change \(sin^3x\) into \[\frac{3sin(x)-sin(3x)}{4}\]
Then, it is *almost* the same as finding the nth derivative of sin(x)

Answer 28

mmk. using that "yahoo" way of showing it, it would work.

Answer 29

LOL! Using @.Sam. 's way too :P
http://openstudy.com/study#/updates/4f74445ce4b0b478589d9a0c

Answer 30

It's basically using the power rule for sin(x).