f(x) = 1/(x+1)+ln{(5x+5)/(5x+6)}

To differentiate f(x).

Solution:

{u(x)+v(x)}' = u'(x) +v'(x).

We use { u (v(x) ) }' = u'(v)* v'(x).

{u(x)/v(x)}' = {u'(x)v(x)-u(x)v'(x)}/[v(x)]^2.

Therefore,

f'(x) = (1/(x+1))' + {ln[(5x+5)/(5x+6)]}'

= -1/(x+1)^2 + [(5x+6)/(5x+5)] {(5x+5)/5x+6)}'

= -1/(x+1)^2 + [(5x+6)/5x+5)] { [(5x+5)'(5x+6) - (5x+5)(5x+6)]/(5x+6)^2

= -1/(x+1)^2 +[(5x+6)/(5x+5)] { [5(5x+6) - (5x+5)5]/(5x+6)^2}

= -1/(x+1)^2 +[(5x+6)/(5x+5)]{25x+30-25x-25)}/(5x+6)^2

=-1/(x+1)^2 +5/{5x+5)(5x+6)}

=-1/(x+1)^2 +1/(x+1)(5x+6)

= (1/(x+1)) {-1/(x+1)-1/(5x+6)}

= [1/(x+1)^2]{-5x-6+(x+1)}/(5x+6)

= -(4x+5)/{(x+1)^2(5x+6)}

We'll re-write the expression using the quotient rule of logarithms:

Supposing that the function is:

f(x)=1/(x+1) + ln[(5x+5)/(5x+6)]

f(x)=1/(x+1) + ln(5x+5) - ln (5x+6)

We'll apply the rule of the derivative of the quotient:

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

**f'(x)= -1/(x+1)^2 + 5/(5x+5) - 5/(5x+6)**