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Math Help - trouble finding derivatives

  1. #1
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    trouble finding derivatives

    I am having trouble solving these problems, and finding the derivatives. I do not need them solved, but just explain the next step (unless otherwise noted):

    1) sin(2x)*cos(2x)
    ' = sin(2x)(-2sin2x)+cos2x(2cos(2x))
    - where did the two come from? wouldn't it just be cos(2x)?

    2) cot(x)/sin(x) = cos(x)/sin^2(x)
    - shouldn't there be a (-) since in the identity, there is one?

    3) 4sec^2(x)
    - Alright, this one, I do not quite understand what they did. It looks like they took the derivative of the number (they get 8) times the sec, then times the derivative of sec^2.
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  2. #2
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    Re: trouble finding derivatives

    Quote Originally Posted by droidus View Post
    I am having trouble solving these problems, and finding the derivatives. I do not need them solved, but just explain the next step (unless otherwise noted):
    1) sin(2x)*cos(2x)
    ' = sin(2x)(-2sin2x)+cos2x(2cos(2x))
    - where did the two come from? wouldn't it just be cos(2x)?
    You are using the chain rule. Differentiate \sin() then 2x.
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  3. #3
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    Re: trouble finding derivatives

    Quote Originally Posted by droidus View Post
    I am having trouble solving these problems, and finding the derivatives. I do not need them solved, but just explain the next step (unless otherwise noted):

    1) sin(2x)*cos(2x)
    ' = sin(2x)(-2sin2x)+cos2x(2cos(2x))
    - where did the two come from? wouldn't it just be cos(2x)?
    The same place the "2" in the first term came from- the "2x" in the trig function. The derivative of sin(u), with respect to u, is cos(u) and if u= 2x, du/dx= 2.
    \frac{d(sin(2x)}{dx}= \frac{d(sin(u))}{du}\frac{du}{dx}= cos(u)(2)= 2cos(2x)

    2) cot(x)/sin(x) = cos(x)/sin^2(x)
    - shouldn't there be a (-) since in the identity, there is one?
    No, why should there be? cot(x)= cos(x)/sin(x) and that sin(x) already in the denominator gives cos(x)/sin^2(x).
    (You may be thinking of the "-" in the derivative of cot(x). There is NO derivative here, just an identity. If you are asking about the derivative of cot(x)/sin(x), you haven't differentiated yet. What does the quotient rule give for the derivative of cos(x)/sin^2(x)?

    3) 4sec^2(x)
    - Alright, this one, I do not quite understand what they did. It looks like they took the derivative of the number (they get 8) times the sec, then times the derivative of sec^2.
    No, they did not take the derivative of any "number", they used the chain rule. Letting u= sec(x), this function is 4u^2, which has derivative, with respect to u, of 4(2u)= 8u= 8 sec(u). Now multiply that by the derivative of u with respect to x, the derivative of sec(x).
    Last edited by HallsofIvy; May 30th 2012 at 12:10 PM.
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  4. #4
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    Re: trouble finding derivatives

    oh, ok - see for the first one... thanks!
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