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A formal power series $f={\sum }_{n\ge 0}{a}_n{x}^n$ is a unit exactly when its constant coefficient is a unit in $A$. If $fg=1$, constant coefficients give $a_0{b}_0=1$. Conversely, for invertible $a_0$, setThen every positive-degree coefficient of $fg$ is zero and its constant coefficient is one, so $g=\sum b_n{x}^n$ is an inverse. This is a formal coefficient calculation and requires no convergence.

Since $A$ is a local ring, its nonunits are exactly $\mathfrak{m}$. The nonunits of $B$ are consequently the series with $a_0\in \mathfrak{m}$, which are exactly $I=\mathfrak{m}B+xB$: such a series is $a_0+xg$, and conversely an element of $I$ has constant coefficient in $\mathfrak{m}$. The map $B\to k$ taking the constant coefficient modulo $\mathfrak{m}$ is surjective with kernel $I$, so $I$ is a maximal ideal. Every proper ideal consists of nonunits and hence lies in $I$, proving that $I$ is the unique maximal ideal. ThusThis proves the description of formal power series over a local ring.

To compute the cotangent space of a formal power series ring, expandDefine the $k$-linear mapChanging a lift $a$ by ${\mathfrak{m}}^2$ or a lift $b$ by $\mathfrak{m}$ changes the image by $I^2$, so the map is well-defined. Every series in $I$ is congruent modulo $x^{2}B\subseteq I^2$ to its first two coefficients, proving surjectivity. If $a+bx\in I^2$, its constant coefficient lies in ${\mathfrak{m}}^2$ and its linear coefficient lies in $\mathfrak{m}$, so both input classes are zero. This proves injectivity. ThereforeFinally, part (a) gives generators $a_1,\dots ,a_d$ of $\mathfrak{m}$, so $I=(a_1,\dots ,a_d,x)$. Use the Krull height theorem in its lecture form: in a Noetherian ring, a prime minimal over an ideal generated by $r$ elements has height at most $r$. The problem allows Noetherianity of $B$ to be assumed. Apply the theorem to $I$ itself, whose height is the Krull dimension of the local ring $B$. Hence $\boxed{\dim B=\mathrm{ht}(I)\le d+1}$.