Using power series prove that $\frac{d} {dx} e^{2x} = 2xe^{2x}$

Question

This is what I have worked on so far:

Since
$$e^x=\sum_{n=0}^\infty \frac{x^n}{n!},$$ then $$e^{2x}=\sum_{n=0}^\infty \frac{2^nx^n}{n!}$$ and $$2xe^{2x}=\sum_{n=0}^\infty \frac{2^{n+1}x^{n+1}}{n!}.$$

Then I take the derivative of $e^{2x}$: $$\frac{d} {dx} \sum_{n=0}^\infty \frac{2^nx^n}{n!}=\sum_{n=0}^\infty \frac{2^n}{n!}nx^{n-1}=\sum_{n=0}^\infty \frac{2^nnx^{n-1}}{n(n-1)!}=\sum_{n=0}^\infty \frac{2^nx^{n-1}}{(n-1)!}$$

Now I don't know how to make $$\sum_{n=0}^\infty \frac{2^nx^{n-1}}{(n-1)!}$$ equal to $$2xe^{2x}=\sum_{n=0}^\infty \frac{2^{n+1}x^{n+1}}{n!}.$$

Answer 1

You can't prove what is false: if $f(x)=e^{2x}$, then $f'(x)=2e^{2x}$, not $2xe^{2x}$. And what you did proves that.

Answer 2

Starting from

$$e^{2x}=\sum_{n=0}^\infty \frac{2^nx^n}{n!}$$

you differentiated term-by-term to form

$$\frac{d}{dx}(e^{2x})=\sum_{{n=1}}^\infty \frac{2^nx^{n-1}}{(n-1)!}=\sum_{n=0}^\infty \frac{2^{n+1}x^{n}}{n!}=2\sum_{n=0}^\infty \frac{2^nx^n}{n!}=2e^{2x}$$

which shows that $\frac{d}{dx}(e^{2x})=2e^{2x}$.