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Use part 1 of the Fundamental Theorem of Calculus to find the derivative.

  1. Jan 13, 2013 #1
    1. h(x) = ∫-3 to sin(x) of (cos(t^3) + t)dt


    2. Okay, I know that you are supposed to replace t with the upper limit, and then I think you multiply that term by the derivative of the upper limit.

    So I thought it would be:

    cos(sinx)^3 * cos(x) + sinxcosx

    But what even is cos(sinx)? I've seen sinxcosx before but not that...
    But I put this answer anyway into my computer hw system, and it's counted wrong.

    Would you please tell me what I'm doing wrong?
    Thank you! :)
     
  2. jcsd
  3. Jan 13, 2013 #2

    Dick

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    (cos(sin(x)^3)+sin(x))*cos(x) is not the same thing as ((cos(sin(x))^3+sin(x))*cos(x). Do you see the difference? Which would you say is correct? I suspect you aren't using enough parentheses.
     
    Last edited: Jan 13, 2013
  4. Jan 13, 2013 #3

    Dick

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    I'll try to use some amateurish tex here to try to emphasize the difference between the important parts. [itex]\cos((\sin x)^3)[/itex] vs [itex](\cos(\sin x))^3[/itex] Which one do you want?
     
  5. Jan 13, 2013 #4
    You were absolutely right! The parentheses were off!

    cos((sinx)^3) should be the right one. I only want to cube the sinx.

    The right answer was:

    [cos(sin(x)^3) + sin(x)]cos(x)

    Oh, this may sound dumb, but... if you were to multiply this out, how would it be written?

    Could you condense the cosines (to something like cos^2(x)(sin(x)^3) or would it have to remain as cos(sin(x)^3)*cos(x) ?

    cos^2(x) means cos(x)*cos(x)... But there are not two x's... So is cos(sin(x)^3)*cos(x) the most simplified version?

    Thank you so much for helping! :)
     
  6. Jan 13, 2013 #5

    Dick

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    cos(sin(x)^3) is the most simplified version. There's not much you can do with it. There's no trig law that will let you expand it out.
     
    Last edited: Jan 13, 2013
  7. Jan 13, 2013 #6
    Oh, okay. Thank you so much! :)
     
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