Maclaurin Series Question

Hi, I'm working on a question about Maclaurin Series:

Show that, if you assume that a function can be differentiated by differentiating each term of its Maclaurin expanion, you get the correct results for
a) d/dx e^x b) d/dx sinx c) d/dx ln(1+x) d) d/dx sqrt(1+x)

For part a, I have tried to use the standard result for e^x and differentiate it, giving 1+x+(1/2)x^2+... but d/dx e^x is e^x so I don't really understand what I'm meant to do next.
Any help would be appreciated, thanks.

Hi, I'm working on a question about Maclaurin Series:

Show that, if you assume that a function can be differentiated by differentiating each term of its Maclaurin expanion, you get the correct results for
a) d/dx e^x b) d/dx sinx c) d/dx ln(1+x) d) d/dx sqrt(1+x)

For part a, I have tried to use the standard result for e^x and differentiate it, giving 1+x+(1/2)x^2+... but d/dx e^x is e^x so I don't really understand what I'm meant to do next.
Any help would be appreciated, thanks.

Just checking: is this A-level or university?

A-Level Further Maths

When you differentiate the series expansion of e^x, you get the exact same series expansion, which proves that d/dx(e^x) = e^x. So you have done the question lol

When you differentiate the series expansion of e^x, you get the exact same series expansion, which proves that d/dx(e^x) = e^x. So you have done the question lol

Maybe I'm being picky, but it only verifies the result. Doesn't prove it. Unless u derive the expansion without differentiating.

Strictly speaking, the question says "show you get the correct result" (and isn't terribly clear what it means (*)). It doesn't ask you to prove d/dx(e^x) = e^x.

(*) What I would think is expected is:

(a) Find (which may mean "look up" ) the Maclaurin expansion. i.e. $f(x) = \sum a_n x^n$, where we have a formula for a_n in terms of n.
(b) Differentiate f(x) and then find the Maclaurin expansion for f'(x), i.e. $f'(x) = \sum b_n x^n$.
(c) Differentiate the series in (a) term by term to get $f'(x) = \sum n a_n x^{n-1}$
(d) Verify that these series are the same. That is, (comparing coefficients of x^n), $n a_n = b_{n-1}$ for $n \neq 0$.

At A-level, it's probably fine to do it for the first 3 terms or so, an university I'd expect you to do it for all n.

At the same time - it's worth noting that it's actually tautologically true that the series have to match up: by definition

$a_n = \dfrac{f^{(n)}(0)}{n!}$ and $b_n = \dfrac{g^{(n)}(0)}{n!}$, where g(x) = f'(x). So in fact $b_n = \dfrac{f^{(n+1)}(0)}{n!} = (n+1)a_{n+1}$.