Differentiate $y=(sin(x))^(log(x))$ ?

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Answer 1 · 2018-01-02T06:47:15

$(dy)/(dx)=0.4343(sinx)^(logx)(sinx/x+cotxlnx)$

As $y=(sin(x))^(log(x))$

we have $lny=ln(sin(x))^(log(x))=logxln(sinx)=lnx/ln10ln(sinx)$

Now as $lny=lnx/ln10ln(sinx)$

$1/y(dy)/(dx)=1/ln10(1/xln(sinx)+lnx*1/sinx*cosx)$

= $1/2.3026(sinx/x+cotxlnx)$

or $(dy)/(dx)=1/2.3026(sinx/x+cotxlnx)xx(sinx)^(logx)$

= $0.4343(sinx)^(logx)(sinx/x+cotxlnx)$

Differentiate y=(sin(x))^(log(x))y=(sin(x))^(log(x))?