Now, let's take the first derivative of $f(x)$:
$f'(x) = m(x-c)^{m-1}g(x) + (x-c)^m g'(x)$.
Notice that $(x-c)$ is still a factor of $f'(x)$. We can continue taking derivatives until we reach the $(m-1)$th derivative:
$f^{(m-1)}(x) = \text{polynomial in }(x-c)
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