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Question

Differentiate the function with respect to x:
xsinx+(sinx)cosx

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Solution

Let y=xsinx+(sinx)cosx

Taking Log both sides
logy=sinxlogx+cosxlog(sinx)

Now Differentiate wrt x
1ydydx=ddx(sinxlogx)+ddx(cosxlog(sinx))1ydydx=(sinx.1x+cosx.logx)+(cosx.1sinx.(cosx)sinxlogsinx)dydx=[sinx.1x+cosx.logx+cos2xsinxsinxlogsinx][xsinx+(sinx)cosx]

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