$n$-th derivative of $\frac{e^x−1}{x}$

Question

Consider the function $g(x) = \frac{e^x-1}{x}$. Find a general formula for $g^{(n)}(x)$ (i.e. the $n-th$ derivative of $g(x)$) and prove that this formula is correct.

• I just want to ask can I use, finding the sum of a series by differentiating or should I use they Taylor series

Answer 1

Hint: $\exp(x) =1+x+x^2/2!+...$ and the Taylor expansion of a function $g(x)$ evaluated at $x=0$ is given by

$$g(x)=\sum_{n=0}^{\infty}\dfrac {g^{(n)}(0)}{n!} x^n.$$

Answer 2

Here is a closed form involving only finite sum. Write $g$ as

$$ g(x) = \int_{0}^{1} e^{xs} \, \mathrm{d}s. $$

Then for $n \geq 1$, we have

\begin{align*} \frac{x^n}{n!}g^{(n)}(x) &= \int_{0}^{1} \frac{x^n s^n}{n!} e^{xs} \, \mathrm{d}s \\ &= \left[ \frac{x^{n-1} s^n}{n!} e^{xs} \right]_{s=0}^{s=1} - \int_{0}^{1} \frac{x^{n-1}s^{n-1}}{(n-1)!} e^{xs} \, \mathrm{d}s \\ &= \frac{x^{n-1}}{n!}e^x - \frac{x^{n-1}}{(n-1)!}g^{(n-1)}(x). \end{align*}

Solving this recurrence relation yields

\begin{align*} g^{(n)}(x) = (-1)^n \frac{n!}{x^{n+1}} \left( e^x \left(\sum_{k=0}^{n} (-1)^{k} \frac{x^k}{k!}\right) - 1 \right) \end{align*}

We remark that this is also easily derived from Leibniz product rule.