Question

Help:see attachment

Answer 1

\[\lim_{h \rightarrow 0} \frac{ 4(x + h)^{2} - 4x ^{2} }{ h } = \lim_{h \rightarrow 0} \frac{ 4x ^{2} + 8xh + 4h ^{2} - 4x ^{2} }{ h }\]

Answer 2

ok I am following just one question where did the 8 come from in the second formula

Answer 3

\[\lim_{h \rightarrow 0} \frac{ 8xh + 4h ^{2} }{ h } = \lim_{h \rightarrow 0} (8x + 4h) = 8x\]

Answer 4

4 times 2xh

Answer 5

ok I gotcha know

Answer 6

so when you got 8x is that the final answer cause I got a little lost

Answer 7

Yes, all you really need is my first 2 posts and 8x is the end of the 2nd post.

Answer 8

ok so then just trying to make sure are you saying the first post the second equation is the derivative or 8x is the derivative

Answer 9

8x is the derivative of 4x^2

Answer 10

oh ok I gotcha know I am following

Answer 11

So that's it. It comes from:\[f'(x) = \lim_{h \rightarrow 0} \frac{ f(x + h) - f(x) }{ h }\]

Answer 12

ok thanks I get a littel ocnfused with the derivatives so this helps thanks

Answer 13

uw! Bazinga!