Question

Question #f3d4d

Answer 1

See below.

Explanation:

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)`