The general nth derivative formula of $(1-x/4)^{-2}$ The question: Determine the nth general (derivative) formula for the function, $$f(x) = \frac{1}{(1-\frac{x}{4})^2}$$

My approach:


Determine $f(x), f'(x) ... f^4 $
Compare the patterns and determine a general formula.


$$f'(x) = \frac{-32}{(x-4)^3}$$
$$f''(x) = \frac{96}{(x-4)^4}$$
$$f'''(x) = \frac{-384}{(x-4)^5}$$

My thoughts: It seems as the the next numerator is a result of the previous numerator multiplied by $ (n+1)*(-1)^{(n+1)}, ex: 96 = -32 * 3, -384 = 96 * -4 $ and the denominator is simply $(x-4)^{n+1}$

I have tried a few different attempts as well but can't seem to complete my logic. This question is meant to help me solve a question regarding the Maclaurin series (where I am suppose to find the series using the definition of the Maclaurin series.
$$f(x) = \frac{1}{(1-\frac{x}{4})^2}=(1-\frac {x}{4})^{-2}$$, $$f(0)=1$$

$$f'(x)= -2(-\frac{1}{4})(1-\frac{x}{4})^{-3} $$ $$f'(0)=\frac{2!}{4}$$  

$$f''(x) = \frac{3!}{4^2}(1-\frac{x}{4})^{-4} $$  $$f''(0)=\frac{3!}{4^2}$$
$$f'''(x) = \frac{4!}{4^3}(1-\frac{x}{4})^{-5} $$  $$f'''(0)=\frac{4!}{4^3}$$

Thus $$f^{(n)}(x) = \frac{(n+1)!}{4^n}(1-\frac{x}{4})^{-(n+2)} $$
The higher derivatives at $x=0$ are  $$f^{(n)}(0) = \frac{(n+1)!}{4^n} $$
You are very close. $f(x)$ can be rewritten as:

$$f(x) = \frac{16}{(4-x)^2}$$

So, the factors of the derivative of order $n$ are:

$$16 \times (-1)^n \times (n+1)!$$

Edit: Hence the nth order derivative is:
$$f^{(n)} = 16 (-1)^n (n+1)! (x-4)^{-2-n}$$

Or, if we switch the order of the terms in the denominator, we can get rid of the $(-1)^n$ term:
$$f^{(n)} = 16 (n+1)! (4-x)^{-2-n}$$