Discuss the continuity and differentiability of $e^x$?

Question

Discuss the continuity and differentiability of $e^x$ ?

---

I know its continuous at all points, but how to check for differentiability ?

Answer 1

This answer assumes that $e^x \equiv \exp x$ is defined with the power series: $$ \exp x = \sum_{j=0}^\infty \frac{x^j}{j!} $$ Using ratio test, $\exp$ converges everywhere: $$ \frac{x^j j!}{x^{j+1} (j+1)!} \to 0 $$ Then we use these two facts:

1. A power series converge compactly within its disk of convergence.

2. If $(f_j)_{j=0}^\infty$ are differentiable functions on $]a,b[$ and $(f_j(x_0))_{j=0}^\infty$ converges for some $x_0 \in \left]a,b\right[$, and $(f_j')_{j=0}^\infty$ converges uniformly on $]a,b[$, then $f_j \to f$ uniformly for some $f$, and $$f'(x) = \lim_{j \to \infty} f_j'(x)$$

The partial sums $\sigma_m(x) = \sum_{j=0}^{m-1} x^j/j!$ converge uniformly on any pre-compact open set within the disk of convergence (a.k.a. $\mathbb{C}$) of $\exp$, and $\sigma_m$s are differentiable as polynomials. Moreover, $\sigma_m(0) \to \exp(0)$. Hence we can invoke (2) to show that $\exp$ is differentiable with derivative given by the limit of $\sigma_m'$.

Answer 2

Partial A.

For any $k$ and for any $d\ne 0$ we have $$(e^{k+d}-e^k)/d=e^k (e^d-1)/d=e^k(e^d-e^0)/(d-0).$$ Fixing $k$ and letting $d\to 0,$ then $e^k$ is fixed and non-zero. So with $f(x)=e^x,$ we see that $f'(k)$ exists iff $f'(0)$ exists.

Now it depends on which def'n of $e$ you are using, to show that $f'(0)=1.$ (Any property of $e$ that no other number has, could be taken as a def'n of $e$).