# Thread: Hard question - nth derivative

1. ## Hard question - nth derivative

$\displaystyle f(x)=\dfrac{x^n}{1-x}$ where n is a constant. Find an expression for the $\displaystyle n^{th}$ derivative, $\displaystyle f^{(n)}(x)$

Thanks!

2. $\displaystyle \displaystyle\frac{x^n}{1-x}=\frac{x^n-1+1}{1-x}=-\frac{1-x^n}{1-x}+\frac{1}{1-x}=$
$\displaystyle =-(1+x+x^2+\ldots +x^{n-1})+\frac{1}{1-x}$

Now, $\displaystyle (1+x+x^2+\ldots+x^{n-1})^{(n)}=0$
and $\displaystyle \left(\frac{1}{1-x}\right)^{(n)}=\frac{(-1)^{n+1}n!}{(1-x)^{n+1}}$

3. Originally Posted by red_dog
$\displaystyle \displaystyle\frac{x^n}{1-x}=\frac{x^n-1+1}{1-x}=-\frac{1-x^n}{1-x}+\frac{1}{1-x}=$
$\displaystyle =-(1+x+x^2+\ldots +x^{n-1})+\frac{1}{1-x}$

Now, $\displaystyle (1+x+x^2+\ldots+x^{n-1})^{(n)}=0$
and $\displaystyle \left(\frac{1}{1-x}\right)^{(n)}=\frac{(-1)^{n+1}n!}{(1-x)^{n+1}}$
From here:
Now, $\displaystyle (1+x+x^2+\ldots+x^{n-1})^{(n)}=0$
and $\displaystyle \left(\frac{1}{1-x}\right)^{(n)}=\frac{(-1)^{n+1}n!}{(1-x)^{n+1}}$

could you briefly explain what you did?

4. Let $\displaystyle g(x)=1+x+x^2+\ldots+x^{n-1}$.
g is a n-1-degree plynomial and the n-th derivative is 0.

Let $\displaystyle h(x)=\frac{1}{1-x}$.
We have $\displaystyle h'(x)=\frac{1}{(1-x)^2}$
$\displaystyle h''(x)=-\frac{2}{(1-x)^3}$
$\displaystyle h'''(x)=\frac{3!}{(1-x)^4}$
By induction $\displaystyle h^{(n)}=\frac{(-1)^{n+1}n!}{(1-x)^{n+1}}$