derivative (1/2)[(1/2)(ln x+1 / x-1 )+arctanx]

Answer 1

we find the derivative (1/2)[(1/2)(ln x+1 / x-1 )+arctanx]

detailed solution given below:

Answer 2

 

Possible derivation:
d/dx(1/2 (1/2 (log(x)+1/x-1)+tan^(-1)(x)))
Factor out constants:
  =  1/2 (d/dx(tan^(-1)(x)+1/2 (-1+1/x+log(x))))
Differentiate the sum term by term and factor out constants:
  =  1/2d/dx(tan^(-1)(x))+(d/dx(-1+1/x+log(x)))/(2)
The derivative of tan^(-1)(x) is 1/(x^2+1):
  =  1/2 (1/2 (d/dx(-1+1/x+log(x)))+1/(x^2+1))
Differentiate the sum term by term:
  =  1/2 (1/(1+x^2)+1/2d/dx(-1)+d/dx(1/x)+d/dx(log(x)))
The derivative of -1 is zero:
  =  1/2 (1/(1+x^2)+1/2 (d/dx(1/x)+d/dx(log(x))+0))
Simplify the expression:
  =  1/2 (1/(1+x^2)+1/2 (d/dx(1/x)+d/dx(log(x))))
Use the power rule, d/dx(x^n) = n x^(n-1), where n = -1: d/dx(1/x) = d/dx(x^(-1)) = -x^(-2):
  =  1/2 (1/(1+x^2)+1/2 (d/dx(log(x))+(-1)/(x^2)))
The derivative of log(x) is 1/x:
  =  1/2 (1/(1+x^2)+1/2 (-1/x^2+1/x))
Simplify the expression:
Answer: |  
 |   =  1/2 (1/2 (-1/x^2+1/x)+1/(1+x^2))