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
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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
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