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

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

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

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Applying the Chain rule,

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

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

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

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

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

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