So as per the matrix cookbook,
$\frac{d |\textbf{A}|}{d \alpha} = |\textbf{A}| \operatorname{Tr}(\textbf{A}^{-1} \frac{d \textbf{A}}{d \alpha})$
where $\textbf{A}$ is a matrix and $\alpha$ is a scalar.
and
$\frac{d |\textbf{A}|}{d \textbf{A}} = |\textbf{A}| (\textbf{A}^{-1})^{T}$
but $\frac{d |\textbf{A}|}{d \alpha}$ can be written as $\frac{d |\textbf{A}|}{d \textbf{A}}.\frac{d \textbf{A}}{d \alpha} = |\textbf{A}| (\textbf{A}^{-1})^{T} \frac{d \textbf{A}}{d \alpha}$ as per the chain rule.
Clearly I have misunderstood something. Why is the result I get with the chain rule not correct?
Is it wrong to apply the chain rule here?
The derivative of the mapping $B \mapsto \det B$ is given by $D \det(B)(H) = \det B \operatorname{tr} (B^{-1} H)$ (assuming that $B$ is invertible).
Hence the chain rule applied to $\det \circ A$ will give $D (\det \circ A) (\alpha) (h) = D \det (A(\alpha)) (D A(\alpha) h) = \det(A(\alpha)) \operatorname{tr} (A(\alpha)^{-1}D A(\alpha) h)$, which is what you have above.