Answer
See the explanation
Work Step by Step
Part (a)
$f(x)=\sum_{n=0}^\infty \frac{x^n}{n!}$
$f(x)=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\ldots$........(1)
Differentiating both sides,
$f'(x)=0+1+\frac{2x}{2!}+\frac{3x^2}{3!}+\frac{4x^3}{4!}+\ldots$
$f'(x)=1+x+\frac{x^2}{2}+\frac{x^3}{3!}+\ldots$.........(2)
Comparing (1) an (2), we conclude $f'(x)=f(x)$.
Part (b)
Define $g(x)=e^x$.
The n-th derivative at $x=0$ is $g^{(n)}(0)=e^0=1$
Then,
$g(x)=\sum_{n=0}^\infty \frac{g^{(n)}(0)x^n}{n!}=\sum_{n=0}^\infty \frac{x^n}{n!}$
Thus, $f(x)=g(x)=e^x$.
