If the function $f(x) = \begin{cases} \frac{(e^{kx}-1) \sin kx}{4 \tan x}, & x \neq 0 \\ P, & x=0 \end{cases}$ is differentiable at $x=0$,then

  • A
    $P=0$,$f^{\prime}(0)=\frac{k^2}{4}$
  • B
    $P=0$,$f^{\prime}(0)=-\frac{1}{2}$
  • C
    $P=k$,$f^{\prime}(0)=-\frac{k^2}{4}$
  • D
    $P=k$,$f^{\prime}(0)=-\frac{1}{4}$

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