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)#