Differentiability proof of exponential function $\sum_{n=0}^ \infty \frac{x^n}{n!}$

Question

$$f(x)=\sum_{n=0}^ \infty \frac{x^n}{n!}$$

I want to prove that $f$ differentiable on $x$ in $[0,1]$.

I am not clear with using the definition of differentiability.

I can prove it is differentiable at $x=0$. I cannot prove in over the interval or when $x$

is something else.

Answer 1

By the very definition (you may be required to provide justifications):

$$f'(x_0)=\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}=\lim_{x\to x_0}\frac1{x-x_0}\sum_{n=1}^\infty\frac{x^n-x_0^n}{n!}=$$

$$=\lim_{x\to x_0}\sum_{n=1}^\infty\frac{x^{n-1}+x_0x^{n-2}+\ldots+x_0^{n-2}x+x_0^{n-1}}{n!}=\sum_{n=1}^\infty\frac{nx_0^{n-1}}{n!}=\sum_{n=0}^\infty\frac{x_0^n}{n!}=f(x_0)$$

Answer 2

Hint:$\displaystyle\sum_{n=0}^{\infty}\dfrac{x^n}{n!}=e^x $ ($e^x$ Taylor series)