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Math Help - Find f'(x)...

  1. #1
    Lil
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    Find f'(x)...

    Find f'(x_0), when f'(x)=2sin(2x)+3cosx+\frac{\pi }{2}, x_{0}=\frac{\pi}{6}
    _______
    I think...
    f'(x)=(2sin(2x)+3cosx+\frac{\pi }{2})'=(2sin(2x))'+(3cosx)'+(\frac{\pi}{2})'=
    =2*cos2x-3sinx=...?
    Last edited by Lil; October 19th 2010 at 07:31 AM.
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  2. #2
    Lil
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    I'll find f'(\frac{\pi}{6}), but i don't know how get first expression
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  3. #3
    Senior Member Sambit's Avatar
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    f'(x) = (4cos2x-3sinx)
    then, f'(x_{0}) = f'(\frac{\pi}{6}) = 4cos2*\frac{pi}{6} - 3sin\frac{pi}{6} = 4*0.5 - 3*0.5 = 0.5
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  4. #4
    Senior Member Sambit's Avatar
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    what is given to you? f(x) or f'(x)?. and if you have to find f'(x) ,why are you trying to determine f(x)?
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  5. #5
    Lil
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    I need find f'(x).

    how u get 4cos2x
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  6. #6
    Senior Member Sambit's Avatar
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    because when you are differentiating sin2x, you are getting 2cos2x, since the derivative of 2x in sin2x is 2.
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