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let f(x)= 3x^3...what is f(a+h)−f(a)/h
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\[\frac{3(a+h)^3-3a^3}{h}\]
so your main job is to cube \[(a+h)^3=a^3+3a^2h+3ah^2+h^3\]
so you get \[\frac{3(a^3+3a^2h+3ah^2+h^3)-3a^3}{h}\] you good from there?
3a^2 +9a^2h + 9ah^2-3a^ over h ??
yes, then note that \[a^3-a^3=0\] and then cancel and h from each term
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\[\frac{9a^2h+9ah^2+3h^3}{h}\] \[9a^2+9ah+3h^2\]
thanks for the help
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