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

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

Find $f'(x_0)$, when $f'(x)=2sin(2x)+3cosx+\frac{\pi }{2},$ $x_{0}=\frac{\pi}{6}$
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I think...
$f'(x)=(2sin(2x)+3cosx+\frac{\pi }{2})'=(2sin(2x))'+(3cosx)'+(\frac{\pi}{2})'=$
$=2*cos2x-3sinx=...$?

2. I'll find $f'(\frac{\pi}{6})$, but i don't know how get first expression

3. $f'(x) = (4cos2x-3sinx)$
then, $f'(x_{0}) = f'(\frac{\pi}{6}) = 4cos2*\frac{pi}{6} - 3sin\frac{pi}{6} = 4*0.5 - 3*0.5 = 0.5$

4. what is given to you? $f(x) or f'(x)?$. and if you have to find $f'(x)$,why are you trying to determine $f(x)$?

5. I need find f'(x).

how u get $4cos2x$

6. because when you are differentiating $sin2x$, you are getting $2cos2x$, since the derivative of $2x$ in $sin2x$ is $2$.