Question #f3d4d

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Answer 1 · 2017-03-21T18:07:03

See below.

Calling $y = y_1+y_2$ with

$y_1=(logx)^x$

and

$y_2=(sin^-x)^sinx$

we have

$logy_1=x log(logx)$ and

$(dy_1)/(y_1) = log(logx)dx + dx/log(x)$ and

$(dy_1)/(dx)=y_1(log(log(x))+1/log(x))=(logx)^x(log(log(x))+1/log(x))$

The same procedure for $y_2$

$log(y_2)=sinx log(sin^-1 x)$ then

$(dy_2)/(dx)=y_2(cosxlog(sin^-1x)+(sinx)/(log(sin^-1x)sqrt(1-x^2)))$

or

$(dy_2)/(dx)=(sin^-x)^sinx(cosxlog(sin^-1x)+(sinx)/(log(sin^-1x)sqrt(1-x^2)))$

and after that

$(dy)/(dx)=(dy_1)/(dx)+(dy_2)/(dx)$