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Differentiate (sinx)^(logx) with respect...

Differentiate `(sinx)^(logx)` with respect to `x` :

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Let `y=(sinx)^(logx)`.
Taking logarithm on both sides of (i), we get
`logy=(logx)(log sin x).`
On differentiating both sides of (ii) w.r.t. x, we get
`(1)/(y).(dy)/(dx)=(logx).(d)/(dx)(log sinx)+(logsinx).(d)/(dx)(logx)`
`=log(x).(1)/(sinx).cos x+(log sin x).(1)/(x)`
`=(logx)cotx+((log sin x))/(x)`
`rArr" "(dy)/(dx)=y.[(logx)cotx+((log sin x))/(x)]`
`rArr" "(dy)/(dx)=(sinx)^(logx).[(logx)cotx+((log sinx))/(x)].`
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