= Solution
The <generalized Leibniz rule> follows by <mathematical induction>. The case $n=0$ is the identity $uv=uv$. Suppose the formula holds for $n$ and differentiate every term using the <product rule>:
$$
(uv)^{(n+1)}=\sum_{r=0}^n\binom nr\left[u^{(r+1)}v^{(n-r)}+u^{(r)}v^{(n-r+1)}\right].
$$
For an interior index $r$, the coefficient of $u^{(r)}v^{(n+1-r)}$ becomes $\binom n{r-1}+\binom nr=\binom{n+1}r$ by <Pascal's identity>. The two endpoint coefficients are one. This proves
$$
\boxed{(uv)^{(n)}=\sum_{r=0}^n\binom nr u^{(r)}v^{(n-r)},\qquad\binom nr=\frac{n!}{r!(n-r)!}.}
$$
Only finitely many <derivatives> are involved; it suffices that both <functions> are $n$ times differentiable.
For $u=x(x+1)$ and $v=e^{2x}$, the <derivatives> of $u$ beyond order two vanish. The three surviving terms of the <generalized Leibniz rule> give
$$
\begin{aligned}
\frac{d^n}{dx^n}\bigl[x(x+1)e^{2x}\bigr]
&=e^{2x}\left[2^n(x^2+x)+n2^{n-1}(2x+1)+n(n-1)2^{n-2}\right]\\
&=\boxed{2^ne^{2x}\left[x^2+(n+1)x+\frac{n(n+1)}4\right].}
\end{aligned}
$$
The simplified expression works for every integer $n\geq0$, including the two cases where one or more of the three terms are absent.
For $u=x^2$ and $v=\sin x$, use $d^m(\sin x)/dx^m=\sin(x+m\pi/2)$. Again only three terms survive. With $T=x+n\pi/2$, they are
$$
\frac{d^n}{dx^n}(x^2\sin x)=x^2\sin T+2nx\sin(T-\pi/2)+n(n-1)\sin(T-\pi).
$$
Since $\sin(T-\pi/2)=-\cos T$ and $\sin(T-\pi)=-\sin T$, this becomes
$$
\boxed{\frac{d^n}{dx^n}(x^2\sin x)=\bigl[x^2-n(n-1)\bigr]\sin\left(x+\frac{n\pi}2\right)-2nx\cos\left(x+\frac{n\pi}2\right).}
$$
The factors $n$ and $n(n-1)$ remove the nonexistent terms when $n=0$ or $1$.
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