# find the derivative of function at given point

Find $f^{(2012)}(\pi/6)$ if $f(x)=\sin x$

Here's the hint from the question paper: You may use Maclaurin series of $\sin x$ and $\cos x$; the formula $\sin(a+b) = \sin a \cos b + \cos a \sin b$ may be useful.

It is quite complicated to me. Btw this is my math graded homework. I know how to find derivative when $x = 0$, but i can't proceed with this qn coz of the $\pi/6$. Any ideas?

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The hint makes it look much harder than it really is. Look at @LJym89’s hint below instead. – Brian M. Scott Oct 28 '12 at 3:21
I see, i see. Thanks for pointing that out. So it is not difficult at all. – uohzxela Oct 28 '12 at 3:30

Notice $f(x) = \sin x$, $f'(x) = \cos x$, $f''(x) = - \sin x$, $f'''(x) = - \cos x$ and $f^{(IV)}(x) = \sin x$. How can you use this information to find the nth derivative of $f$ ? This is a hint.

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2012 % 4 = 0. So the derivative is sinx? So the result is sin($\pi$/6) = 1/2? It can't be that easy? Why is there a need for the complicated hint sin(a+b)?? – uohzxela Oct 28 '12 at 3:25
@uohzxela Sometimes people, even sharp mathematicians, go for pretty messy stuff obviating simpler approaches. Read about the two trains puzzle and von Neumann's answer to it..and the astonishing way he obtained it here: mathworld.wolfram.com/TwoTrainsPuzzle.html – DonAntonio Oct 28 '12 at 4:31

$$Let \space y=sin(x)$$

$$y'=cos(x)=sin(x+\pi/2)$$

$$y''=-sin(x)=sin(x+2(\pi/2))$$

$$y'''=-cos(x)=sin(x+3(\pi/2))$$

$$...$$

$$y^{(n)}(x)=sin(x+n(\pi/2))$$

You could now simplify this expression and your specific value: finding-the-values-of-cos-fracn-pi2-and-sin-fracn-pi2.

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