1. **Problem statement:** Find the $n^{th}$ derivative of the function $f(x) = \tan(2x)$.\n\n2. **Recall the function and derivative rules:** The function is $f(x) = \tan(2x)$. The derivative of $\tan(u)$ with respect to $x$ is $\sec^2(u) \cdot \frac{du}{dx}$. Here, $u = 2x$, so $\frac{du}{dx} = 2$.\n\n3. **First derivative:**\n$$f'(x) = \frac{d}{dx} \tan(2x) = \sec^2(2x) \cdot 2 = 2 \sec^2(2x)$$\n\n4. **Pattern for higher derivatives:** The derivatives of $\tan(x)$ involve powers of $\sec(x)$ and $\tan(x)$, and can be expressed using the formula for the $n^{th}$ derivative of $\tan(x)$, but with the chain rule applied for $2x$.\n\n5. **General formula:** The $n^{th}$ derivative of $\tan(2x)$ can be written as\n$$f^{(n)}(x) = 2^n \cdot \frac{d^n}{du^n} \tan(u) \bigg|_{u=2x}$$\nwhere $\frac{d^n}{du^n} \tan(u)$ is the $n^{th}$ derivative of $\tan(u)$ with respect to $u$.\n\n6. **Summary:** To find $f^{(n)}(x)$, compute the $n^{th}$ derivative of $\tan(u)$ with respect to $u$, then multiply by $2^n$ and substitute $u=2x$.\n\n**Final answer:**\n$$f^{(n)}(x) = 2^n \cdot \frac{d^n}{du^n} \tan(u) \bigg|_{u=2x}$$