Consider the function $f(x)=e^x$.
a. Differentiate the Taylor series about 0 of $f(x)$.
b. Identify the function represented by the differentiated series.
c. Give the interval of convergence of the power series for the derivative.

Step 1 ：The Taylor series of a function $f(x)$ about $x=a$ is given by the formula: $$f(x)=f(a)+f^{\prime }(a)(x-a)+\frac{f^{″}(a)}{2!}(x-a{)}^2+\frac{f^{‴}(a)}{3!}(x-a{)}^3+\cdots$$

Step 2 ：For the function $f(x)=e^x$, the derivative of $e^x$ is also $e^x$, and $e^0=1$, so the Taylor series of $f(x)$ about $x=0$ is: $$f(x)=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$$

Step 3 ：Differentiating this series, we get: $$f^{\prime }(x)=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$$

Step 4 ：This is the same as the original series, so the function represented by the differentiated series is also $f(x)=e^x$.

Step 5 ：The power series for $e^x$ converges for all $x$, so the interval of convergence of the power series for the derivative is also all real numbers.

Step 6 ：Final Answer: $$\boxed{{\displaystyle \begin{array}{rl}a.& \text{ The differentiated Taylor series about 0 of }f(x)=e^x\text{ is }{f}^{\prime }(x)=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots .\\ b.& \text{ The function represented by the differentiated series is }f(x)=e^x.\\ c.& \text{ The interval of convergence of the power series for the derivative is all real numbers.}\end{array}}}$$