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Perhaps a simple derivative?

  1. Apr 5, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the derivative of the function: q = sin ([itex]\frac{t}{\sqrt{t+1}}[/itex])

    Answer: cos ([itex]\frac{t}{\sqrt{t+1}}[/itex]) ([itex]\frac{t+2}{2(t+1)^{\frac{3}{2}}}[/itex])

    2. Relevant equations

    Chain Rule
    [itex]\frac{dq}{dt}[/itex] sin x = cos x

    3. The attempt at a solution

    [itex]\frac{dq}{dt}[/itex] = cos ([itex]\frac{t}{\sqrt{t+1}}[/itex]) [itex]\frac{dq}{dt}[/itex] (t(t+1))[itex]^{-\frac{1}{2}}[/itex] = cos ([itex]\frac{t}{\sqrt{t+1}}[/itex]) (t(-[itex]\frac{1}{2}[/itex](t+1)[itex]^{-\frac{3}{2}}[/itex] + 1(t+1)[itex]^{-\frac{1}{2}}[/itex])

    So that's as far as I've gotten with this problem. I unfortunately don't know how to continue with it though. Does simplifying the derivative of (t(t+1))[itex]^{-\frac{1}{2}}[/itex] lead me to the answer provided? Or did I derive something wrong?
     
  2. jcsd
  3. Apr 5, 2012 #2
    Yes, you're right simplification will lead you to the right answer.

    so you have

    -t/(2(t+1)^(3/2)) + 1/(t+1)^(1/2)

    so the common denominator is 2(t+1)^(3/2) so multiply the top and bottom of the second expression by 2(t+1)

    so you have

    -t/(2(t+1)^(3/2)) + 2(t+1)/(2(t+1)^(3/2))

    add them now

    2t+2-t/(2(t+1)^(3/2))

    t+2/(2(t+1)^(3/2))
     
  4. Apr 5, 2012 #3
    Ah thank you! I've figured it out now and learned a new thing about exponents.
     
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