I was trying to calculate the $n^{th}$ derivative of the function$$f(x) = \dfrac{1}{1-x}$$
And I already found a couple of derivatives but I'm not able to find the pattern.
$f_1(x) = \dfrac{1}{(1-x)^{2}}$
$f_2(x) = \dfrac{2}{(1-x)^{3}}$
$f_3(x) = \dfrac{6}{(1-x)^{5}}$
$f_4(x) = \dfrac{30}{(1-x)^{9}}$
$f_5(x) = \dfrac{270}{(1-x)^{17}}$
Can someone please help me find the pattern in this derivative?
$\endgroup$ 41 Answer
$\begingroup$The derivative of $x^{n}$ is $nx^{n-1}$, whatever integer is $n$. Thus the derivative of $(1-x)^n$ is $-n(1-x)^{n-1}$ and the sequence you get is therefore$$ \frac{1}{1-x}\quad \frac{1}{(1-x)^2}\quad \frac{2}{(1-x)^3}\quad \frac{6}{(1-x)^4}\quad \cdots $$and a simple induction will confirm it: the $n$-th derivative is$$ f^{(n)}(x)=\frac{n!}{(1-x)^{n+1}} $$Indeed, the derivative of the function $n!(1-x)^{-n-1}$ is$$ n!(-n-1)(-1)(1-x)^{-n-2}=\frac{(n+1)!}{(1-x)^{n+2}} $$
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