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

Jun 10, 2018

$y ' = 12 \cos \left(2 t\right) - 4 \sin \left(4 t\right)$

#### Explanation:

Using that

$\left(\sin \left(x\right)\right) ' = \cos \left(x\right)$
$\left(\cos \left(x\right)\right) ' = - \sin \left(x\right)$
and
$\left(u + v\right) ' = u ' + v '$
and the chain rule we get

$y ' = 6 \cos \left(2 t\right) \cdot 2 - \sin \left(4 t\right) \cdot 4$
so
$y ' = 12 \cos \left(2 t\right) - 4 \sin \left(4 t\right)$

Jun 10, 2018

$y ' = 4 \left(3 \cos \left(2 t\right) - \sin \left(4 t\right)\right)$

#### Explanation:

Let me know if you need clarification on any steps.

Jun 10, 2018

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

#### Explanation:

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

Applying the Chain rule,

$y ' = 6 \cdot d \left(\sin \left(2 t\right)\right) + d \left(\cos \left(4 t\right)\right)$

$y ' = 6 \cos 2 t \cdot d \left(2 t\right) - \sin 4 t \cdot d \left(4 t\right)$

$y ' = 6 \cdot \cos 2 t \cdot 2 - \sin 4 t \cdot 4$

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

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