= Solution
The <inverse different> is the <trace-dual lattice>
$$
\mathfrak D_{L/K}^{-1}=\{a\in L:\operatorname{Tr}_{L/K}(a\mathcal O_L)\subseteq\mathcal O_K\}.
$$
The integral closure $\mathcal O_L$ is finite free over the complete <discrete valuation ring> $\mathcal O_K$. Choose an integral basis and use the nondegenerate <trace pairing> to form its dual basis over $K$. This exhibits $\mathfrak D^{-1}$ as a full, finite $\mathcal O_K$-lattice. It is stable under multiplication by $\mathcal O_L$, because $b\mathcal O_L\subseteq\mathcal O_L$ for $b\in\mathcal O_L$, and bounded denominators make it a <fractional ideal> of $\mathcal O_L$. Integral elements have integral <field traces>, so $\mathcal O_L\subseteq\mathfrak D^{-1}$. A nonzero <fractional ideal> of a <discrete valuation ring> is invertible; its inverse is consequently an integral <ideal>, the <different ideal> $\mathfrak D_{L/K}$.
Assume now $A=\mathcal O_K[x]=\mathcal O_L$, with monic separable <minimal polynomial> $g$ of degree $n$. <Lagrange interpolation> at its distinct roots gives
$$
\operatorname{Tr}_{L/K}\!\left(\frac{x^j}{g'(x)}\right)=\begin{cases}0&0\leq j<n-1,\\1&j=n-1.\end{cases}
$$
Indeed, these are the leading coefficients in the interpolation formula for $X^j$. For any element of $A$, this trace is the coefficient of $X^{n-1}$ in its degree-less-than-$n$ representative modulo $g$. The resulting pairing on the basis $1,x,\ldots,x^{n-1}$ is integral and unimodular: reversing the order of one basis makes its matrix triangular with diagonal ones, since entries vanish when the exponent sum is less than $n-1$. Thus it identifies $A$ with its full $\mathcal O_K$-dual. Translating back to the <trace pairing> proves
$$
\boxed{\mathfrak D_{L/K}^{-1}=g'(x)^{-1}\mathcal O_L,\qquad\mathfrak D_{L/K}=g'(x)\mathcal O_L.}
$$
For a <totally ramified extension> of degree $n$, any <uniformizer> $\pi_L$ is an <Eisenstein generator of a totally ramified extension>, and $\mathcal O_L=\mathcal O_K[\pi_L]$. To see the latter equality, use the common <residue field> to expand an integral element in powers of $\pi_L$ with digits from $\mathcal O_K$; reduce powers using its <Eisenstein polynomial> and take limits in the finite complete $\mathcal O_K$-module generated by $1,\pi_L,\ldots,\pi_L^{n-1}$. Write that polynomial as $X^n+\sum_{j<n}a_jX^j$, with $v_L(a_j)\geq n$ for $j<n$. When $p\nmid n$, the derivative's leading term has <valuation> $n-1$, while every other nonzero derivative term has <valuation> at least $n$. There can be no cancellation of the unique smallest term. Therefore
$$
\boxed{\mathfrak D_{L/K}=\pi_L^{n-1}\mathcal O_L.}
$$
For the prime-power <p-adic cyclotomic extension>, put $\zeta=\zeta_{p^r}$ and $e=p^{r-1}(p-1)$. Modulo $p$, the shifted <cyclotomic polynomial> $\Phi_{p^r}(1+T)$ is $T^e$, while its constant term is $p$, not a multiple of $p^2$. It is therefore <Eisenstein>, so $\pi=\zeta-1$ is a <uniformizer> and $\mathcal O_{K_r}=\mathbb Z_p[\zeta]$. From
$$
\Phi_{p^r}(X)=\frac{X^{p^r}-1}{X^{p^{r-1}}-1}
$$
we obtain, by differentiating the numerator at $\zeta$,
$$
\Phi'_{p^r}(\zeta)=\frac{p^r\zeta^{-1}}{\zeta^{p^{r-1}}-1}.
$$
The denominator is a primitive-$p$ <root of unity> minus one and has $K_r$-<valuation> $p^{r-1}$. The <different exponent> is consequently $r e-p^{r-1}$, giving
$$
\boxed{\mathfrak D_{K_r/\mathbb Q_p}=(\zeta-1)^{p^{r-1}(r(p-1)-1)}\mathcal O_{K_r}.}
$$
This includes $p=2,r=1$: the extension is trivial and its <different exponent> is zero.
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