Let
\[
f(x)=e^{2x}\sin(3x).
\]
Using Leibniz’s formula directly, we obtain a binomial sum first, then collapse it into the same amplitude–phase expression. The general Leibniz rule is
\[
(uv)^{(n)}=\sum_{k=0}^{n}\binom{n}{k}u^{(n-k)}v^{(k)}.
\]
 [en.wikipedia](https://en.wikipedia.org/wiki/General_Leibniz_rule)

## Apply Leibniz’s formula

Choose

\[
u(x)=e^{2x},
\qquad
v(x)=\sin(3x).
\]

The relevant derivatives are

\[
u^{(m)}(x)=2^m e^{2x},
\]

and, because derivatives of sine cycle through sine/cosine with a phase shift,

\[
v^{(k)}(x)=3^k\sin\left(3x+\frac{k\pi}{2}\right).
\]

Therefore,

\[
\begin{aligned}
f^{(n)}(x)
&=
\sum_{k=0}^{n}
\binom{n}{k}
u^{(n-k)}(x)v^{(k)}(x)\\[4pt]
&=
\sum_{k=0}^{n}
\binom{n}{k}
\left(2^{n-k}e^{2x}\right)
\left[
3^k\sin\left(3x+\frac{k\pi}{2}\right)
\right].
\end{aligned}
\]

Factor out \(e^{2x}\):

\[
\boxed{
f^{(n)}(x)
=
e^{2x}
\sum_{k=0}^{n}
\binom{n}{k}
2^{n-k}3^k
\sin\left(3x+\frac{k\pi}{2}\right)
}.
\]

This is the \(n\)-th derivative in a form obtained **purely by Leibniz’s formula**.

## Separate sine and cosine

Using

\[
\sin\left(3x+\frac{k\pi}{2}\right)
=
\sin(3x)\cos\left(\frac{k\pi}{2}\right)
+
\cos(3x)\sin\left(\frac{k\pi}{2}\right),
\]

we get

\[
\begin{aligned}
f^{(n)}(x)
={}&e^{2x}
\Bigg[
\sin(3x)
\sum_{k=0}^{n}
\binom{n}{k}
2^{n-k}3^k
\cos\left(\frac{k\pi}{2}\right)
\\
&\qquad+
\cos(3x)
\sum_{k=0}^{n}
\binom{n}{k}
2^{n-k}3^k
\sin\left(\frac{k\pi}{2}\right)
\Bigg].
\end{aligned}
\]

So in the usual form,

\[
f^{(n)}(x)=e^{2x}\bigl(A_n\sin(3x)+B_n\cos(3x)\bigr),
\]

where

\[
\boxed{
A_n=
\sum_{k=0}^{n}
\binom{n}{k}
2^{n-k}3^k
\cos\left(\frac{k\pi}{2}\right)
}
\]

and

\[
\boxed{
B_n=
\sum_{k=0}^{n}
\binom{n}{k}
2^{n-k}3^k
\sin\left(\frac{k\pi}{2}\right)
}.
\]

## Evaluate the binomial sums

Let

\[
\theta=\tan^{-1}\left(\frac32\right).
\]

Then

\[
2+3i=\sqrt{13}\,(\cos\theta+i\sin\theta).
\]

The binomial theorem gives

\[
(2+3i)^n
=
\sum_{k=0}^{n}
\binom{n}{k}2^{n-k}(3i)^k.
\]

Its real and imaginary parts are precisely the two sums above:

\[
A_n=\Re((2+3i)^n),
\qquad
B_n=\Im((2+3i)^n).
\]

By polar form,

\[
(2+3i)^n
=
(\sqrt{13})^n
\bigl(\cos(n\theta)+i\sin(n\theta)\bigr).
\]

Hence

\[
A_n=13^{n/2}\cos(n\theta),
\qquad
B_n=13^{n/2}\sin(n\theta).
\]

Substitute this into the Leibniz result:

\[
\begin{aligned}
f^{(n)}(x)
&=
e^{2x}13^{n/2}
\left[
\cos(n\theta)\sin(3x)
+
\sin(n\theta)\cos(3x)
\right]\\[4pt]
&=
13^{n/2}e^{2x}\sin(3x+n\theta).
\end{aligned}
\]

Thus,

\[
\boxed{
\frac{d^n}{dx^n}\left(e^{2x}\sin(3x)\right)
=
13^{n/2}e^{2x}
\sin\left(3x+n\tan^{-1}\frac32\right)
}.
\]

## Check for \(n=2\)

From the Leibniz sum,

\[
\begin{aligned}
f''(x)
={}&
\binom20(2^2e^{2x})\sin(3x)
+
\binom21(2e^{2x})(3\cos 3x)
\\
&+
\binom22e^{2x}(-9\sin 3x).
\end{aligned}
\]

Therefore,

\[
f''(x)
=
e^{2x}\left(4\sin 3x+12\cos 3x-9\sin 3x\right),
\]

so

\[
\boxed{
f''(x)=e^{2x}\left(-5\sin(3x)+12\cos(3x)\right)
}.
\]

The closed formula yields the same result, since \(13^{1}=13\) and the angle is \(2\tan^{-1}(3/2)\).