If $f(x)=e^x$,$g(x)=\sin^{-1} x$ and $h(x)=f(g(x))$,then $\frac{h^{\prime}(x)}{h(x)}$ is

  • A
    $e^{\sin^{-1} x}$
  • B
    $\frac{1}{\sqrt{1-x^2}}$
  • C
    $\sin^{-1} x$
  • D
    $\frac{e^{\sin^{-1} x}}{\sqrt{1-x^2}}$

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