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Thread: Find f'(x)...

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

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

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