Let $f(x)=(x-a)^{n}g(x)$ such that $g(a)\neq 0$; compute its $n$-th derivative

Question

Let $f(x)=(x-a)^ng(x)$ such that $g(a)\neq 0$; then I would like to compute the $n$-th derivative of $f$:

$$f^{(n)}(x)=\dfrac{d^n}{dx^{n}}\Bigl((x-a)^n\cdot g(x) \Bigr)$$

Can I have any suggestion?

Answer 1

You can use the generalized rule for derivative of a product of two functions:

$$(fg)^{(n)}=\sum_{k=0}^{n}{n\choose k}f^{(n-k)}g^{(k)}$$

For more functions is like the multinomial theorem. Also, when $n>r$ you will have:

$$ \frac{d^{r}}{dx^{r}}(x-a)^{n} = \frac{n!}{(n-r)!}(x-a)^{n-r}$$

Answer 2

Hint:

By Leibnitz formula we have

$$f^{(n)}(a)= ((x-a)^ng(x))^{(n)}=\sum_{k=0}^{n}{n\choose k}((x-a)^n)^{(k)}g^{(n-k)}(x)|_{x=a}$$

But $$((x-a)^n)^{(k)}|_{x=a} =0~~~~~k\neq n$$ can you continue from here?