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

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

Solution:

f(x) = )=(x+1)^-1 - ln(x+1)/(x+2).

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

f'(x) = {1/(x+1)}'- {ln(x+1)/(x+2)}'.

f'(x) = -1/(x+1)^2 - {(x+2)/(x+1)}{(x+1/(x+2)}, as {ln u(x)}' = {1/u(x)}u'(x).

f'(x) = -1/(x+1)^2 + (x+2)/(x+1){(x+1)'(x+2)-(x+1)(x+2)'}/(x+2)^2, as {f(x)/g(x) = {f'(x)g(x)- f(x)g'(x)}/{g(x)}^2.

f'(x) = -1/(x+1)^2 -{x+2)/(x+1){(x+2 -x-1}/(x+2)^2.

f'(x) = -1/(x+1)^2 - (x+2)/{(x+1)(x+2)^2}.

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

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

**Therefore ****for 0< x< infinity**** f'(x) = -(x+3)/{(x+1)(x+2)}.**

We'll re-write the expression of the function, applying the rule of negative power for the first term and the quotient rule of logarithms, for the 2nd term.

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

Now, we'll calculate the first derivative of f(x):

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

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

For the first term, we'll differentiate using the quotient rule:

(u/v)'=(u'*v-u*v')/v^2

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

[ln(x+1)]' = 1/(x+1) (3)

[ln(x+2)]' = 1/(x+2) (4)

We'll substitute (2), (3), (4) in (1):

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

f'(x) = [-(x+2) - (x+1)(x+2) + (x+1)^2]/(x+2)*(x+1)^2

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

We'll combine and eliminate like terms:

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

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