Question 295296
First, let's compute {{{f(a+h)}}}, we do this by simply replacing each 'x' with 'a+h' and simplifying. So {{{f(a+h)=2/(a+h+1)}}}


Similarly, doing the same for {{{f(a)}}} gets us {{{f(a)=2/(a+1)}}}



This then means that {{{f(a+h)-f(a)=2/(a+h+1)-2/(a+1)}}}



So the goal now is to simplify {{{2/(a+h+1)-2/(a+1)}}}



Take note that the LCD is {{{(a+h+1)(a+1)}}}. So multiply the first fraction by {{{(a+1)/(a+1)}}} to get {{{(2(a+1))/((a+h+1)(a+1))-2/(a+1)}}}



Now multiply the second fraction by {{{(a+h+1)/(a+h+1)}}} to get {{{(2(a+1))/((a+h+1)(a+1))-(2(a+h+1))/((a+h+1)(a+1))}}}



Combine the fractions: {{{(2(a+1)-2(a+h+1))/((a+h+1)(a+1))}}}



Distribute: {{{(2a+2-2a-2h-2)/((a+h+1)(a+1))}}}



And finally combine like terms: {{{(-2h)/((a+h+1)(a+1))}}}



So in the end, {{{f(a+h)-f(a)=(-2h)/((a+h+1)(a+1))}}}