Let $\text{SU}(2)$ denote the special group of $2 \times 2$ unitary matrices, that is, unitary matrices with determinant $1$.
Define $f : \text{SU}(2) \times \text{SU}(2) \to \text{SU}(2) \times \text{SU}(2)$ by
$$f(A, B) = (BAB^{-1}, B)$$
Let $A, B \in \text{SU}(2)$ be such that $A$ and $B$ commute.Find $\mathrm{d}f_{(A, B)}$.
Here's what I have so far (I'm pretty sure the solution is valid, but would like a second opinion / critiques):
By definition, $\mathrm{d}f_{(A, B)}$ is the map taking $(C, D)$ to $$ \lim_{t\to 0} \frac{f((A, B) + t(C, D)) - f(A, B)}{t}. $$ Then $$ (B + tD)(A + tC)(B + tD)^{-1} = (B + tD)(A + tC)(B^* + tD^*) = A + tBAD^* + tBCB^* + t^2 BCD^* + tDAB^* + t^2 DAD^* + t^2 DCB^* + t^3 DCD^* $$ where the first equality comes from the fact that the matrices are unitary. So the numerator in the limit is $(tBAD^* + tBCB^* + t^2 BCD^* + tDAB^* + t^2 DAD^* + t^2 DCB^* + t^3 DCD^*, tD)$ and after dividing this by $t$ we get $$ df_{(A, B)}(C, D) = \lim_{t\to 0} (BAD^* + BCB^* + t BCD^* + DAB^* + t DAD^* + t DCB^* + t^2 DCD^*, D) = (BAD^* + BCB^* + DAB^*, D) $$
The first equality is wrong; $B^{-1}=B^*$ and $D^{-1}=D^*$ don't imply $(B+tD)^{-1}=B^*+tD^*$. You need to expand the inverse in powers of $t$:
\begin{align} (B+tD)^{-1}&=\left(B\left(1+tB^{-1}D\right)\right)^{-1}\\ &=\left(1+tB^{-1}D\right)^{-1}B^{-1}\\ &=\left(1-tB^{-1}D+O\left(t^2\right)\right)B^{-1}\\ &=B^*-tB^*DB^*+O\left(t^2\right)\;. \end{align}
So
\begin{align} (B + tD)(A + tC)(B + tD)^{-1}&=(B + tD)(A + tC)(B^*-tB^*DB^*)+O\left(t^2\right)\\ &=BAB^*+t(DAB^*+BCB^*-BAB^*DB^*)+O\left(t^2\right)\\ &=BAB^*+t(DAB^*+BCB^*-ADB^*)+O\left(t^2\right)\\ &=BAB^*+t(DA+BC-AD)B^*+O\left(t^2\right)\;, \end{align}
where the penultimate step uses the fact that $A$ and $B$ commute. Thus, the differential is $((DA+BC-AD)B^*,D)$.