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

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

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