Yes, well done! Personally I think you should write everything out fully all the time so that
you know exactly what you're doing & can see the logic to all of it.
I'll just show you what I'm talking aboutf(r \ + \ h) \ - \ f(r)} \ = \ \frac{ar^2 \ - \ ar^2 \ - \ 2arh \ - ah^2}{\pi r^2 (r + h)^2}f(r \ + \ h) \ - \ f(r)} \ = \ \frac{- \ 2arh \ - ah^2}{\pi r^2 (r + h)^2}
\frac{f(r \ + \ h) \ - \ f(r)}{h} \ = \ \frac{- \ 2arh \ - ah^2}{ h \pi r^2 (r + h)^2}
\frac{f(r \ + \ h) \ - \ f(r)}{h} \ = \ \frac{- \ 2ar \ - ah}{ \pi r^2 (r + h)^2}
You see, everything is logically laid out so that you can't make a mistake, try to get things
written out this way until you can do all of this in your head.
All you have to do is to take the limit of both sides now
\lim_{h \to 0} \ \frac{f(r \ + \ h) \ - \ f(r)}{h} \ = \ \lim_{h \to 0} \ \frac{- \ 2ar \ - ah}{ \pi r^2 (r + h)^2}
See! On the left hand side you have the definition of the derivative clearly shown!
On the right you need to send h to 0 & look, a term dissappears & the h in the denominator
also dissappears so that you get a clean answer.
\lim_{h \to 0} \ \frac{f(r \ + \ h) \ - \ f(r)}{h} \ = \ \lim_{h \to 0} \ \frac{- \ 2ar}{ \pi r^2 r^2}
\lim_{h \to 0} \ \frac{f(r \ + \ h) \ - \ f(r)}{h} \ = \ \frac{- \ 2ar}{ \pi r^4}
\lim_{h \to 0} \ \frac{f(r \ + \ h) \ - \ f(r)}{h} \ = \ \frac{- \ 2a}{ \pi r^3}
a = 12.75
\lim_{h \to 0} \ \frac{f(r \ + \ h) \ - \ f(r)}{h} \ = \ \frac{- \ 25.5}{ \pi r^3}Check out the videos, they'll give you a lot of good explanations for all of this & I do advise
getting that cheap book, it's better than Stewart, Thomas etc...
to keep an eye on what you're 
but please tell me why \pi is in the numerator?
