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find the n th derivative of y=sinx and hence find the 9th derivative of the function.
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\[ \cos(x) = \sin(\pi/2-x) \]
\[ \sin(\pi/2-x) = -\sin(x-\pi/2) \]
It helps to move out the negative, because \(\frac{d}{dx} \pi/2-x = -1\) but \(\frac{d}{dx} x-\pi/2 = 1\)
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