Question 895964
I'm going to assume the function is {{{f(x) = 4/(x+9)}}}


First we need f(a)


{{{f(x) = 4/(x+9)}}}


{{{f(a) = 4/(a+9)}}} Replace every 'x' with 'a'


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Now we need f(a+h)


{{{f(x) = 4/(x+9)}}}


{{{f(a+h) = 4/(a+h+9)}}} Replace every 'x' with "a+h"


{{{f(a+h) = 4/(a+9+h)}}}


{{{f(a+h) = 4/((a+9)+h)}}}


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Now because both f(a) and f(a+h) have an "a+9" in the denominator, I'm going to let z = a+9


So {{{f(a) = 4/(a+9)}}} turns into {{{f(a) = 4/z}}} 


{{{f(a+h) = 4/((a+9)+h)}}} turns into {{{f(a+h) = 4/(z+h)}}}


this will make the work shown below a bit more easier

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Subtract the two. We do so by finding the LCD (in this case z(z+h)) and then getting each fraction to have this LCD


{{{f(a+h) - f(a) = 4/(z+h) - 4/z}}}


{{{f(a+h) - f(a) = (4z)/(z(z+h)) - 4/z}}} Multiply top and bottom of the first fraction by 'z' 


{{{f(a+h) - f(a) = (4z)/(z(z+h)) - (4(z+h))/(z(z+h))}}} Multiply top and bottom of the second fraction by 'z+h' 


{{{f(a+h) - f(a) = (4z)/(z(z+h)) - (4z+4h)/(z(z+h))}}}


{{{f(a+h) - f(a) = (4z - (4z+4h))/(z(z+h))}}}


{{{f(a+h) - f(a) = (4z - 4z-4h)/(z(z+h))}}}


{{{f(a+h) - f(a) = -(4h)/(z(z+h))}}}


{{{f(a+h) - f(a) = -(4h)/((a+9)(a+9+h))}}} Replace every 'z' with 'a+9'


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The last step is to divide all of that by 'h'. Put another way, we multiply all of that by {{{1/h}}}. This is because dividing is the same as multiplying by the reciprocal.


{{{f(a+h) - f(a) = -(4h)/((a+9)(a+9+h))}}}


{{{(1/h)*(f(a+h) - f(a)) = (1/h)*(-(4h)/((a+9)(a+9+h)))}}} Multiply both sides by {{{1/h}}}


{{{(f(a+h) - f(a))/h = (1/highlight(h))*(-(4*highlight(h))/((a+9)(a+9+h)))}}} We have these common 'h' terms (in top and bottom)


{{{(f(a+h) - f(a))/h = (1/cross(h))*(-(4*cross(h))/((a+9)(a+9+h)))}}} which divide to 1 and cancel out (they go away)


{{{(f(a+h) - f(a))/h = -4/((a+9)(a+9+h)))}}}


which is the final answer. You could expand out the denominator, but that makes things messier than they have to be.