Given #y=(sin(x))^(logx)# calculate #dy/dx# ?

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Answer 1 · 2017-04-03T10:05:25

#dy/dx = (sin(x))^(logx)(log(sinx)/x+log xtanx)#

Supposing that the question is about

#y=(sin(x))^(logx)#

Applying #log# to both sides

#logy=log x log(sinx)#

now deriving

#(dy)/y = dx/x log(sinx)+logx cosx/sinx dx#

so

#dy/dx = y(log(sinx)/x+log xtanx)#

and finally

#dy/dx = (sin(x))^(logx)(log(sinx)/x+log xtanx)#