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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)#