# Calculate f'(x), x belongs to ( 0, infinite) if f:(o,infinite)->R, f(x) = 1/(x+1) + ln(2x+1)/(2x+3)

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f(x) = 1/(x+1)+ln((2x+1)/(2x+3))

f(x)=1/(x+1)+ln(2x+1)-ln(2x+1), as ln(a/b) lna-lnb.

f'(x) = (1/(x+1))'+(ln(2x+1)'-(ln(2x+3))'

= -1/(x+1)^2+2/(2x+1)-2/(2x+3),

=-1/(x+1)^2+2{(2x+3-(2x+1))/(2x+1)(2x+3)}

=-1/(x+1)^2+4/[(2x+1)(2x+3)]

=[(2x+3)(2x+1)-4(x+1)^2)]/[(x+1)^2(2x+3)(2x+1)]

= [4x^2+8x+3-4x^2-8x-4]/[(x+1)^2(2x+1)(2x+3)]

=-1/(2x+3)(x+1)^2(2x+1)

= which does not exist when x=-1.5, -1 , and -0.5

f'(x) is negative for x >-0.5,

f'(x) is positive when -1 <x< -0.5

f'(x) is negative when -1.5 <x < -1.

f'(x) positive when x < -1.5

If the expression contains "ln[(2x+1)/(2x+3)]", we know the rule that says that ln(f/g)=ln f -ln g.

Supposing that the function is:

f(x) = 1/(x + 1) - ln[(2x + 1)/(2x + 3)]

f(x) = 1/(x + 1) - ln(2x + 1) + ln (2x + 3)

Applying the rule of the fraction derivative, which says

(f/g)' = (f'*g - f*g')/g^2

f'(x) = -1/(x + 1) - 2/(2x + 1) + 2/(2x + 3)