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)`
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