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Question

Find dydx, if y=xsinx.

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Solution

y=xsinx
logy=sinxlogx
ddx(logy)=ddx(sinxlogx)
1ydydx=sinx×ddx(logx)+logxddx(sinx)
1ydydx=sinx×1x+logxsinx
dydx=y(sinxx+logxsinx)
dydx=xsinx(sinxx+logxsinx)

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