Answer:
[tex]f^{(k)}(x)=\dfrac{17k!(-1)^k}{(x-9)^{k+1}}[/tex]
Step-by-step explanation:
The question presumes you have access to a computer algebra system. The one I have access to provided the output in the attachment. The list at the bottom is the list of the first four derivatives of f(x).
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The derivatives alternate signs, so (-1)^k will be a factor.
The numerators start at 17 and increase by increasing factors: 2, 3, 4, indicating k! will be a factor.
The denominators have a degree that is k+1.
Putting these observations together, we can write an expression for the k-th derivative of f(x):
[tex]\boxed{f^{(k)}(x)=\dfrac{17k!(-1)^k}{(x-9)^{k+1}}}[/tex]
