Find the 100th derivative of $x \sinh(2x)$ If $f(x) = x \sinh(2x)$, find $f^{({100})}(x)$.
My (Incorrect) working so far:
Using Leibniz' Formula for derivatives:
$$(fg)^{(n)}=\sum_{k=0}^n{n\choose k}f^{(k)}g^{(n-k)}$$
$$(x\sinh(2x))^{(100)}=\sum_{k=0}^{100}{100\choose k}f^{(k)}g^{(n-k)}=$$
Since the second derivative of $x$ is $0$...
$$=\sum_{k=0}^1{2\choose k}f^{(k)}g^{(n-k)}$$
$$={2\choose 0}(x)(\sinh(2x))+{2\choose 1}(1)(2\cosh(2x))$$
$$=4x \sinh(2x)+4 \cosh(2x)$$
This is my answer and it is not correct. What do I need to change to make it correct?
Thanks in advance.

I have a better method, perhaps, if your aim is to evaluate this at $x=0$. Of course, other methods are required if you want the derivative $\forall x \in \mathbb{R}$. These have already been discussed in the comments and answers. Regardless this is a good way to 'check' whether you haven't made any computational mistakes along the way!
The Taylor expansion gives $x\cdot \frac{f^{99}(0)\cdot x^{99}}{99!} = \frac{g^{100}(0)\cdot x^{100}}{100!}$ where $f(x) = \sinh(2x)$ and $g(x) = x\cdot \sinh(2x)$.
Now $f^{99}(0) = 2^{99}\cdot \cosh(2\cdot 0) = 2^{99}$ and so $g^{100}(0) = 100\cdot 2^{99}$.
