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Stuck for 1hr on this UGLY derivative: 1/3sinx^3 (don't mock me)

  1. Jan 5, 2006 #1
    It is "sin x to the power of 3 over 3". How do I derive this? I am supposed to use the CHAIN RULE.

    Now the answer I am supposed to get is -sin^2x times cosx.

    How did "that" happen. How do I chain the monster, 1/3sinx^3?

    What troubles me is the sinx^3. For a regular sinx, the derivative is cosx. But with powers...? cosx^3?

    Please 'elp.

    Thx all
     
  2. jcsd
  3. Jan 5, 2006 #2
    if you rewrite it as (sinx)^-3 divided by 3 its easier to se how to use the chain rule. D(F(G(x))=F'(G(x))*G'(x)

    G(x) in this case is sinx
     
  4. Jan 5, 2006 #3
    Az,

    Why would I write (Sinx)^ -3? Then I would just be taking the reciprocal of sinx^3.... which is not what I want right?

    Now I keep seeing THREE parts to the chain process.
    So Mr. Chain tells us to do this... f'(g(x)) times g'(x).

    And you already told me that g(x) is sinx. (therefore g'(x) = cos x)

    So the f(g(x)) part would be 1/3 x^3. I still got more than a single variable in there! Then I would have to CHAIN AGAIN.
    (In this case x^3 for the outside function, and 1/3x for the inside function...)

    What am I not getting?
     
  5. Jan 5, 2006 #4
    No, no, scrap the last bit.

    1/3x^3 bit IS differentiable. but I am still not getting the correct answer. :mad:
     
  6. Jan 5, 2006 #5

    arildno

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    Dearly Missed

    This means that the sine of x appears to the power of 3, not that x is to the power of 3.

    The sign is wrong in the "answer" you gave.
     
  7. Jan 5, 2006 #6
    Thanks guys.

    I still don't quite get it, but I'm not even sure what to ask, so I'll leave it at that.
    Yeah, the answer was "negative sine of x to the power of 2 times cosx". (made a mistake). (-sinx)^2 * cosx for those who would still like to have a crack at this.

    Later all
     
  8. Jan 5, 2006 #7
    I think I got confused by how you used /

    is it 1/(3sinx^3) or one third of sinx^3 ???

    if its the second scrap my suggestion.

    By the answere I take it its the second one but it cant possibly be a negative answere since it starts positive and the derivate of sin is cos.


    F(x)=(x^3)/3 => F'(x)=(x^2)*3*1/3

    Now just derivate G(x)=sinx

    you should get the correct answere from that and there is no way it can become negative. The correct answere should be
    cosx*(sinx)^2
    not
    -(sinx^2) * cosx
     
  9. Jan 5, 2006 #8
    and I apologise had I read your post properly I would have understood its one third of (sinx)^3 :blushing: :yuck:
     
  10. Jan 5, 2006 #9

    Tx

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    3Cosec^2x * Cosecx
    3(1 + Cot^2x) * Cosecx
    3Cosecx + 3Cos^2x/Sin^3x
    3Cosecx + 3Tan^2xCosecx
    3Cosecx(Sec^2x)
    3CosecxSec^2x

    That should be pretty easy to differentiate. IE Product Rule
     
    Last edited by a moderator: Jan 5, 2006
  11. Jan 5, 2006 #10

    HallsofIvy

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    You wrote 1/3sinx^3

    Do you MEAN (1/3) sin3(x)
    or
    (1/3)sin(x3)
    or
    1/(3 sin3(x))
    or
    1/(3 sin(x3))??

    If y= (1/3) sin3(x) then y'= (1/3)(3 sin2(x))(cos(x))= sin2(x)cos(x).

    If y= (1/3)sin(x3) then y'= (1/3)(cos(x3))(3x2= x2cos(x3)

    If y= 1/(3 sin3(x))= (1/3) sin-3(x) then y'= (1/3)(-3 sin-2(x))cos(x)= - cos(x)/sin2(x).

    If y= 1/(3 sin(x3))= (1/3)(sin(x3))-1 then y'= (1/3)(-1)(sin(x3))-2(cos(x3)(3x2= -x2(cos(x3)/(sin(x3))2.
     
  12. Jan 5, 2006 #11

    shmoe

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    You've said the answer was supposed to be "-sin^2x times cosx" and you later said "(-sinx)^2 * cosx". These are different. Please be very careful when transcribing problems into ascii, use extra parenthesis if unsure. See Hall's post above for the many ways your question could be interpreted as you wrote it.
     
  13. Jan 5, 2006 #12
    It is HallsofIvy's FIRST OPTION.

    I am sorry for all the confusion. I finally got it. The step I forgot to do all along was SUBSTITUTE the sinx part INTO the derivative of the OUTER FUNCTION (Using the chain rule).
    I was sitting there with the derivative of the outer function (3/3*x^2) and wondering how the heck there is a sin2(x) in the answer.

    Anyways please bear with me, I will be able to express myself mathematically some day.... (I hope...).

    Thank you everybody for your efforts
     
  14. Jan 28, 2006 #13
    Let me try. been a while so don't laugh.
    1/3(sin(x))^3 i rewrote it for confort.
    carry the power forward. 3/3 becomes 1 so i'll not write it at all.
    sin(x)^2 times the derivative of what's inside. inside is a x so...1*sin(x)^2 is what i'd get.
    let me check. i'll try another spomething. (3x^3+3x^2)^3
    3(3x^3+3x^2)^2*(9x^2+6x)
    so i was right. I have the sensation the "cos" should appear in there but...guess not.

    edit: he said he got it so i wrote the full thing for the heck of it.
     
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