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!

  • 2
    $\begingroup$ What do you understand by "the alternative form of the derivative"? This is not standard terminology. $\endgroup$ – Alex M. Oct 10 '16 at 15:05
  • $\begingroup$ @AlexM. I'd guess he means $\lim_{x \to c}\frac{f(x)-f(c)}{x-c}$ $\endgroup$ – GFauxPas Oct 10 '16 at 15:17

From sources online I can only interpret the 'alternative' form as this:

$$ f^{\prime}(x) = \lim_{x\to c} \frac{f(x)-f(c)}{x-c} $$


$$ \begin{align} f^{\prime}(2) &= \lim_{x\to 2} \frac{x^3 + 4x - (2^3 + 4 \cdot 2))}{x - 2}\\ &= \lim_{x\to 2} \frac{x^3 + 4x - 16}{x - 2}\\ &= \lim_{x\to 2} \frac{(x-2)(x^2 + 2x + 8)}{x-2}\\ &= \lim_{x\to 2} \, (x^2 + 2x + 8)\\ &= 16 \end{align} $$

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?

  • $\begingroup$ Yes I factored it our incorrectly. Thank you so much, 16 is what my graph was giving me as the answer so I appreciate your help! Thanks again! $\endgroup$ – Brianna J Oct 10 '16 at 16:12
  • $\begingroup$ Sorry I meant *out $\endgroup$ – Brianna J Oct 10 '16 at 16:12

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