Question

find the nth derivative of Sin(4x).Cos(2x)
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Answer 1

Answer:

your answer

Step-by-step explanation:

y=sin4x=(1−cos2x2)2=14⋅[1+cos22x−2cos2x]

=14⋅[1+1+cos4x2−2cos2x]

⟹y=18⋅[3+cos4x−4cos2x]

∴y′n=18⋅

Answer 2

$\frac{d}{dx} cos(2x)sin(4x)\\\\=\frac{d}{dx} [cos(2x)].sin(4x)+ cos(2x).\frac{d}{dx} [sin(4x)]\\\\=(-sin(2x)).\frac{d}{dx}[2x].sin(4x) + cos(2x)cos(4x).\frac{d}{dx} [4x] \\\\=-2\frac{d}{dx} [x].sin(2x)sin(4x)+cos(2x)cos(4x).4.\frac{d}{dx} [x]\\\\=-2.1.sin(2x)sin(4x)+4cos(2x)cos(4x).1\\\\=4cos(2x)cos(4x) -2sin(2x)sin(4x)\\\\=3cos(6x)+cos(2x)$

For further reference refer to :

https://brainly.in/question/49292973

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