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Question

If y = xx + (sinx)cotx . find dydx

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Solution

dydx=d(x)xdx+(sinx)cotxdx

Let z=xx

Take log on both sides

logz=xlogx

Now differentiate above equation with respect to x

1z×dzdx =1+logx

dzdx=xx(1+logx)


Let t=(sinx)cotx

Take log on both sides

logt=cotx×log(sinx)
Now differentiate with respect to x

1t×dtdx=cosec2x×log(sinx)+cotx×1sinx×cosx

dtdx= (sinx)cotx×(cosec2x×log(sinx)+cot2x)

dydx=dtdx+dzdx

Put the values of dzdxanddtdx

dydx=xx(1+logx)+(sinx)cotx×(cosec2x×log(sinx)+cot2x)

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