How do you find the fourth derivative of $sin x$ $sin x$ ?

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How do you find the fourth derivative of $sin x$ $sin x$ ?

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Answer 1 · 2016-04-19T14:36:23

$y = sin x$ $y=sinx$
$\Rightarrow y_{1} = \frac{d}{d x} (sin x)$ $=>y_1=d/(dx)(sinx)$
$\Rightarrow y_{1} = cos x$ $=>y_1=cosx$
$\Rightarrow y_{2} = \frac{d}{d x} (cos x)$ $=>y_2=d/(dx)(cosx)$
$\Rightarrow y_{2} = - sin x$ $=>y_2=-sinx$
$\Rightarrow y_{3} = \frac{d}{d x} (- sin x)$ $=>y_3=d/(dx)(-sinx)$
$\Rightarrow y_{3} = - cos x$ $=>y_3=-cosx$
$\Rightarrow y_{4} = \frac{d}{d x} (- cos x)$ $=>y_4=d/(dx)(-cosx)$
$\Rightarrow y_{4} = sin x$ $=>y_4=sinx$

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How do you find the fourth derivative of $sin x$ $sin x$ ?

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