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Question

ddx[3sin(2x+π3)+cos(2x+π3)]=____

A
4cos2x
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B
4sin2x
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C
4sin2x
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D
4cos2x
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Solution

The correct option is B 4sin2x
ddx[3sin(2x+π3)+cos(2x+π3)]
=ddx[232sin(2x+π3)+22cos(2x+π3)]
=ddx2[sin(2x+π3)cosπ6+cos(2x+π3)sinπ6]
=ddx2[sin(2x+π3+π6)]
=ddx2[sin(2x+π2)]
=ddx(2cos2x)
=4sin2x

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