Differentiate `(sinx)^(logx)` w.r.t. x.

1.	Differentiate `(sinx)^(logx)` w.r.t. x.
Answer» 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)].`

Discussion

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