How do you find the derivative of #y=6sin(2t) + cos(4t)#?

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Answer 1 · 2018-06-10T04:39:47

#y'=12cos(2t)-4sin(4t)#

Using that

#(sin(x))'=cos(x)#
#(cos(x))'=-sin(x)#
and
#(u+v)'=u'+v'#
and the chain rule we get

#y'=6cos(2t)*2-sin(4t)*4#
so
#y'=12cos(2t)-4sin(4t)#

Answer 2 · 2018-06-10T04:43:27

#y'=4(3cos(2t)-sin(4t))#

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Let me know if you need clarification on any steps.

Answer 3 · 2018-06-10T04:49:09

#color(blue)(y' = 4 * (3 cos 2t - sin 4t)#

#y = 6 sin 2t + cos 4t#

Applying the Chain rule,

#y' = 6 * d(sin (2t)) + d (cos (4t))#

#y' = 6 cos 2t * d(2t) - sin 4t * d(4t)#

#y' = 6 * cos 2t * 2 - sin 4t * 4#

#y' = 12 cos 2t - 4 sin 4t#

#color(blue)(y' = 4 * (3 cos 2t - sin 4t)#