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Question

If y=logx3+3sin1x+kx2, then find dydx

A
31x+311x2+k(2x)
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B
31x3+311x2+k(2x)
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C
31x311x2+k(2x)
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D
31x+311x2+2x
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Solution

The correct option is B 31x+311x2+k(2x)
Here, y=logx3+3sin1x+kx2

On differentiating we get

dydx=ddx[logx3]+ddx[3sin1x]+ddx[kx2]

=3ddx[logx]+3ddx(sin1x)+kddx(x2)

=31x+311x2+k(2x)

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