This is seriously so long I'm not sure it will even fit on a single line. The formula for the derivative using the limiting process is [tex] \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} [/tex]. And of course this is a problem because if h approaches 0, we would have a 0 in the denominator of that fraction and that is definitely not allowed! Every x in the function will be replaced with (x+h) to give us this: [tex] \lim_{h \to 0} \frac{-(x+h)^3+17(x+h)^2-(x+h)+3}{h} [/tex]. When we expand that we will get this very long numerator (I'm purposely leaving out the limit as h approaches 0 part to save space): [tex] \frac{-(x^3+2x^2h+xh^2+x^2h+2xh^2+h^3)+17x^2+34xh+17h^2-x-h+3-(-x^3+17x^2-x+3)}{h} [/tex]. Simplifying that leaves us with this: [tex] \frac{-x^3-2x^2h-xh^2-x^2h-2xh^2-h^3+17x^2+34xh+17h^2-x-h+3+x^3-17x^2+x-3}{h} [/tex]. We have a lot of terms that cancel each other out so when we do that we are left with [tex] \lim_{h \to 0} \frac{-3x^2h-3xh^2-h^3+34xh+17h^2-h}{h} [/tex]. That is one of your choices for answers, the third one down on the left to be specific. Now we can factor out an h: [tex] \lim_{h \to 0} \frac{h(-3x^2-3xh-h^2+34x+17h-1)}{h} [/tex]. That h on the top outside the parenthesis cancels with the h on the bottom. Now, as h approaches 0 we have no problems! Yay! That means when we now replace h with 0, we have this: [tex] -3x^2-0-0+34x+0-1 [/tex], or simplified we have [tex] -3x^2+34x-1 [/tex] which is also a choice for your answers, top one on the right. Those are your 2 answers for that dertivative. It's much simpler when you learn the rules!
