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what is the derivative of 8sin(4pix)
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you know the derivative of \(\sin(x)\)?
-cos(x)
you know the chain rule?
is it (f(g(x))) = f'(g(x))g'(x)
put your two answers together :)
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-8cos(4pix)
(4pix)
is 8.cos(4.pi.x) if derived to 4.pi.x if derived to x: 8.4.pi.cos (4.pi.x)
close...you need to multiply by the derivative of the inside
yes...plus \[\frac{d}{dx}\sin(x)=\cos(x)\]
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-32cos(x)
and don't forget the \(\pi\)
so the answer is -32cos(pix)
\[32\pi\cos(4\pi x)\]
8sin(4πx) ( sin u )' = u' cos u u = 4πx --> u' = 4π -> 8 * 4π cos (4πx) = 32 π cos (4πx)
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