Use the alternative type of the derivative to find the derivative in ~ $x=c$ if $f(x)=x^3+4x$ and also $c=2$.

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I keep gaining stuck v the answer being $0$, no matter exactly how I try to deal with it. If someone might please usage step-by-step indict to aid me watch what I"m law wrong, that would be great. Thank you!  From sources online I deserve to only analyze the "alternative" kind as this:

$$f^\prime(x) = \lim_x\to c \fracf(x)-f(c)x-c$$

So,

\beginalignf^\prime(2) &= \lim_x\to 2 \fracx^3 + 4x - (2^3 + 4 \cdot 2))x - 2\\&= \lim_x\to 2 \fracx^3 + 4x - 16x - 2\\&= \lim_x\to 2 \frac(x-2)(x^2 + 2x + 8)x-2\\&= \lim_x\to 2 \, (x^2 + 2x + 8)\\&= 16\endalign

The vital step is cancelling the $x-2$ in the denominator, notification we have the right to do this because $x-2\neq 0$ as we space considering the limit. Have the right to you see which point you to be struggling at?

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answer Oct 10 "16 in ~ 15:40 danwalkerdevdanwalkerdev
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