can someone differentiate this

Question

y=√x + loge x-^x/2

Answer 1

Rewrite the first term to x^1/2 Then,

y' = 1/2x^-1/2 + 1/[x ln e]-1/2

Since ln e = 1, this can be simplified

y' = 1/2x^-1/2 + 1/x-1/2