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Derivative of function h(t) = sin(arccos(t)) - help

  1. Oct 4, 2007 #1
    1. The problem statement, all variables and given/known data
    find the derivative of the function:
    h(t) = sin (arccos t)


    2. The attempt at a solution
    derivative of sin = cos
    derivative of arccos t:
    (-1/sqrt 1-t^2) * 1

    my book has a complete different answer so I need some help please.
     
  2. jcsd
  3. Oct 4, 2007 #2
    did u try chain rule?
     
  4. Oct 4, 2007 #3
    Yes I got...
    -1 * (1-t^2)^-1/2 + (2t)
    (cos) * 1/2 * (1-t^2)^-3/2 + (2t)
     
  5. Oct 4, 2007 #4
    no that is incorrect... chain rule will be cos u *du where u in this case arcos(t)..
    try again...
     
    Last edited: Oct 4, 2007
  6. Oct 4, 2007 #5

    Dick

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    No, it won't. BuBbLeS01's first solution was much closer. Derivative of sin(f(t)) is cos(f(t))*f'(t). Let f be arccos.
     
  7. Oct 4, 2007 #6
    hmm I am getting [tex]\frac{-x}{(1-x^2)^(1/2)}[/tex] which I believe is right unless i read the question incorrectly!..
    aaah a typo I have corrected the typo should be cos.thanks for pointing that out!
     
    Last edited: Oct 4, 2007
  8. Oct 4, 2007 #7

    Dick

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    I'll let BuBbLeS01 check that.
     
  9. Oct 5, 2007 #8
    Yes thats right...where did cos go?
     
  10. Oct 5, 2007 #9
    Is this right?
    cos (arccos t) * cos(-1/sqrt 1-t^2) * (2t)
    My teacher does not want our answers simplified at all.
     
  11. Oct 5, 2007 #10

    Dick

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    You are mixing up arccos and it's derivative. cos(arccos(t))=t. cos and arccos are inverse functions, right? The derivative of sin(f(t))=cos(f(t))*f'(t). That's the chain rule. Now carefully put f=arccos.
     
  12. Oct 5, 2007 #11

    arildno

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    Okay, we have:
    1. [tex]h(t)=\sin(arccos(t))[/tex]

    2. [tex]\frac{d}{dx}\sin(x)=\cos(x)[/tex]

    3. [tex]\frac{d}{dt}arccos(t)=\frac{-1}{\sqrt{1-t^{2}}}[/tex]

    4. [tex]\cos(arccos(t))=t[/tex]

    Agreed?

    Thus, by the chainrule, setting [itex]x(t)=arccos(t)[/itex], we have:
    [tex]\frac{dh}{dt}=\cos(x(t))\frac{dx}{dt}=-\cos(arccos(t))\frac{1}{\sqrt{1-t^{2}}}=-\frac{t}{\sqrt{1-t^{2}}}[/tex]
     
  13. Oct 5, 2007 #12
    cos (arccos t) * (-1/sqrt 1-t^2)
     
  14. Oct 5, 2007 #13

    arildno

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    Correct!

    Now, what does cos(arcos(t)) simplify to?
     
  15. Oct 5, 2007 #14
    Ok so cos(arccos(t)) simplifies to 1 because they are inverses of each other. I am sorry it has been a while since I have had precalc so I rusty
     
  16. Oct 5, 2007 #15

    arildno

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    No, it simplifies to..t, not 1!
    They are functional inverses, not multiplicative inverses.
     
  17. Oct 5, 2007 #16
    ok so anytime I have like sin(arcsin(t)) it just cancels to t?
     
  18. Oct 5, 2007 #17

    Dick

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    Sure, just like (sqrt(t))^2=t.
     
  19. Oct 5, 2007 #18
    Okay I have another similar problem...
    Find the derivative of...
    tan(arcsint)
    can I use the product rule?
     
  20. Oct 5, 2007 #19

    Dick

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    It's not a product. Use the chain rule.
     
  21. Oct 5, 2007 #20
    sec^2(arcsint) * (1/sqrt1 + t^2)
     
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