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Math Help - differentiate

  1. #1
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    differentiate

    how do you differentiate 2cos(theta) - 2cos^2(theta)?

    i got -2sin(theta) +2sin^2(theta) but its wrong.....
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  2. #2
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    Re:

    Hi Jeph!!


    Remember that:

    y = cos(x) y' = -sin(x)

    also

    y = sin(x) y' = cos(x)


    the prefix -co in this since means negative

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  3. #3
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    i tohught if you differentiate cos it would turn to sin not cos and sin...

    can you do 2sin(theta) - 2cos(theta)sin(theta) for me?
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  4. #4
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    Re:

    Quote Originally Posted by jeph View Post
    i thought if you differentiate cos it would turn to sin not cos and sin...

    can you do 2sin(theta) - 2cos(theta)sin(theta) for me?
    Sure, but remember...

    Cos(x) does not turn into Sin(x)

    it turns into -Sin(x)

    It's crucial that you put a negative in front of -Sin(x) or else your answer is no good. Also if there was a coefficient in front of Cos(x) such as 4Cos(x) the negative would go out in front thus -4Sin(x).

    You will see some product rule action in this next one:

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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jeph View Post
    i tohught if you differentiate cos it would turn to sin not cos and sin...

    can you do 2sin(theta) - 2cos(theta)sin(theta) for me?
    what happened was he differentiated cos^2(theta), for this you would get sin(theta)cos(theta)

    explanation:
    we have to do the chain rule on this one.

    it might be better to think of cos^2(theta) as (cos(theta))^2. that way, to perform the chain rule, you first have to bring the 2 down and multiply, subtract 1 from the power and leave what's in the brackets, then to complete the chain rule, multiply by the derivative of what's in the brackets.

    y = cos^2(theta)
    => y = (cos(theta))^2
    => y' = 2(cos(theta))*(-sin(theta))
    => y' = -2cos(theta)sin(theta)
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  6. #6
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    oh...hmmmm

    thanks guys. its slowly coming back to me alittle now...
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