Calculus of a Single Variable, Chapter 5, 5.6, Section 5.6, Problem 51

The derivative of y in terms of x is denoted by (dy)/(dx) or y’ .
For the given problem: y = 1/2(1/2ln((x+1)/(x-1)) +arctan(x)) , we may apply the basic differentiation property: d/(dx) c*f(x) = c d/(dx) f(x) .
d/(dx)y =d/(dx) 1/2[1/2ln((x+1)/(x-1)) +arctan(x)]
y'=1/2d/(dx) [1/2ln((x+1)/(x-1)) +arctan(x)]
Apply the basic differentiation property: d/(dx) (u+v) = d/(dx) (u) + d/(dx) (v)
y'=1/2[d/(dx) (1/2ln((x+1)/(x-1))) +d/(dx)(arctan(x))]
For the derivative of d/(dx)(1/2ln((x+1)/(x-1))) , we may apply again the basic derivative property:d/(dx) c*f(x) = c d/(dx) f(x) .
d/(dx) (1/2ln((x+1)/(x-1)))=1/2d/(dx) (ln((x+1)/(x-1)))
For the derivative part, follow the basic derivative formula for natural logarithm function: d/(dx) ln(u)= (du)/u .
Let u =(x+1)/(x-1) then du = -2/(x-1)^2 .
Note For the derivative of u=(x+1)/(x-1) ,we apply the Quotient Rule: d/(dx)(f/g) = (f'*g-f*g')/g^2 .
Let:
f= (x+1) then f'=1
g=(x-1) then g'=1
Then,
d/(dx)((x+1)/(x-1))= (1*(x-1)-(x+1)*(1))/(x-1)^2
=((x-1)-(x+1))/(x-1)^2
=(x-1-x-1)/(x-1)^2
=(-2)/(x-1)^2
Applying: d/(dx) ln(u)= (du)/u on:
1/2d/(dx)(ln((x+1)/(x-1)))= (1/2) *(((-2)/(x-1)^2))/(((x+1)/(x-1)))
=(1/2) *((-2)/(x-1)^2)*(x-1)/(x+1)
=(-2(x-1))/(2(x-1)^2(x+1))
Cancel common factors 2 and (x-1) from top and bottom:
(-2(x-1))/(2(x-1)^2(x+1)) =-1/((x-1)(x+1))
Recall (x-1)*(x+1) = x^2-x+x-1 = x^2-1 then the derivative becomes:
1/2d/(dx)(ln((x+1)/(x-1)))=-1/(x^2-1)

For the derivative of d/(dx)(arctan(x)) , we apply basic derivative formula for inverse tangent:
d/(dx)(arctan(x))=1/(x^2+1)

Combining the results, we get:
y'=1/2[d/(dx) (1/2ln((x+1)/(x-1))) +d/(dx)(arctan(x))]
y'=(1/2) [-1/(x^2-1) +1/(x^2+1)]
y' =(1/2) [-1/(x^2-1) *(x^2+1)/(x^2+1) +1/(x^2+1)*(x^2-1)/(x^2-1)]
y' =(1/2) [(-(x^2+1) +(x^2-1))/((x^2-1) (x^2+1))]
y' =(1/2) [(-x^2-1+x^2-1)/((x^2-1) (x^2+1))]
y' =(1/2) [(-2)/((x^2-1) (x^2+1))]
y' =(-1)/((x^2-1) (x^2+1))
or
y'= (-1)/(x^4-1)