Determine $f^{(5)}(0)$ without computing any derivatives? Determine $f^{(5)}(0)$ without computing any derivatives? How do i approach a problem like this? I think i use maclaurin/taylor series but i'm not completely sure?
$f(x)=x\ln(1+x)$
$f(x)=x\cos x$
 A: $$f(x) = x \log{(1+x)} = \sum_{k=1}^{\infty} (-1)^{k+1} \frac{x^{k+1}}{k}$$
For any sufficiently differentiable function
$$f(x) = \sum_{k=0}^{\infty} \frac{f^{(k)}(0)}{k!} x^k$$
so that
$$\frac{f^{(5)}(0)}{5!} = -\frac{1}{4} \implies f^{(5)}(0) = -30$$
Do a similar thing for 
$$f(x) = x \cos{x} = \sum_{k=0}^{\infty} (-1)^{k} \frac{x^{2 k+1}}{(2 k)!}$$
A: When you have the taylor series 
$$\sum_{k=0}^\infty a_k x^k$$ 
you know the $n$-th derivative is 
$$f^{(k)}(0)=a_k\cdot k! $$
A: You can try and use Leibniz's general product rule:
$$ (fg)^{(n)} = \sum_{r = 0}^{n} \binom{n}{r} f^{(r)} g^{(n-r)}$$
where we use $h^{(m)}$ to denote the $m^{th}$ derivative of $h$.
If $f$ and $g$ have easy derivatives, you can compute the derivative of the product (your second problem).
If $fg$ and $g$ are functions whose derivatives are easily computed, this allows you to give you a recurrence of computing $f^{(k)}(0)$ in terms of $f^{(k-1)}(0), f^{(k-2)}(0)$ and so you don't really need to symbolically compute the whole derivatives of $f$.
This can be useful if you don't know the Taylor series before hand: you get an algorithm to compute it!
For example, consider the generating function for Bernoulli numbers
$$b(z) = \frac{z}{e^z - 1} = \sum_{n=0}^{\infty} B_n \dfrac{z^n}{n!}$$
You can apply the above idea by applying the general product rule to $b(z)(e^z -1)$ to compute the coefficients $B_n$.
