# Calculate f'(x) if 0<x<infinite , f(x)=(x+1)^-1 - ln(x+1)/(x+2).

We have to find f'(x). f(x) is given as (x+1)^-1 - ln(x+1)/(x+2).

[(x+1)^-1 - ln(x+1)/(x+2)]'

we can write ln (x +1)/(x +2) = ln (x+1) - ln(x+2)

=> [(x+1)^-1 - ln (x+1) + ln(x+2)]'

=> [(x+1)^-1]' - [ln (x+1)]' + [ln(x+2)]'

=> -1* (x +1)^-2 - 1/(x +1) + 1/(x+2)

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We have to find f'(x). f(x) is given as (x+1)^-1 - ln(x+1)/(x+2).

[(x+1)^-1 - ln(x+1)/(x+2)]'

we can write ln (x +1)/(x +2) = ln (x+1) - ln(x+2)

=> [(x+1)^-1 - ln (x+1) + ln(x+2)]'

=> [(x+1)^-1]' - [ln (x+1)]' + [ln(x+2)]'

=> -1* (x +1)^-2 - 1/(x +1) + 1/(x+2)

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

now making the denominator common

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

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

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

The required result is:

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

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