If $f(x) = \begin{cases} \frac{x^2-16}{x-4} & \text{if } x > 4 \\ 2x & \text{if } x \leq 4 \end{cases}$ then $f^{\prime}(4^{-}) + f^{\prime}(4^{+}) = $

  • A
    $1$
  • B
    $2$
  • C
    $3$
  • D
    $4$

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