Use the alternative form of the derivative to find the derivative at $x=c$ if $f(x)=x^3+4x$ and $c=2$.
I keep getting stuck with the answer being $0$, no matter how I try to solve it. If someone could please use step-by-step instructions to help me see what I'm doing wrong, that would be great. Thank you!
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$\begingroup$From sources online I can only interpret the 'alternative' form as this:
$$ f^{\prime}(x) = \lim_{x\to c} \frac{f(x)-f(c)}{x-c} $$
So,
$$ \begin{align} f^{\prime}(2) &= \lim_{x\to 2} \frac{x^3 + 4x - (2^3 + 4 \cdot 2))}{x - 2}\\ &= \lim_{x\to 2} \frac{x^3 + 4x - 16}{x - 2}\\ &= \lim_{x\to 2} \frac{(x-2)(x^2 + 2x + 8)}{x-2}\\ &= \lim_{x\to 2} \, (x^2 + 2x + 8)\\ &= 16 \end{align} $$
The important step is cancelling the $x-2$ in the denominator, notice we can do this because $x-2\neq 0$ as we are considering the limit. Can you see which point you were struggling at?
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