Assume $\varphi: [0,1] \rightarrow [0, \infty)$ is continuous, strictly decreasing function.
In the book ([1], p. 91) it is claimed that
\begin{align} & \int_0^1 \int_0^1 \varphi(\varphi^{-1}(u)+\varphi^{-1}(v)) ~d(\varphi(\varphi^{-1}(u)+\varphi^{-1}(v))) \\[10pt] = {} & \int_{0}^{1}\int_{0}^{1} \varphi(\varphi^{-1}(u)+\varphi^{-1}(v)) \cdot \varphi''(\varphi^{-1}(u)+\varphi^{-1}(v)) \cdot [ \varphi' \circ \varphi^{-1}(u)]^{-1} \cdot [ \varphi' \circ \varphi^{-1}(v)]^{-1} \, du \, dv \end{align}
How do we get this result? What are the intermediate steps that were skipped here, if any? If not, what is the general rule for derivation that was used here?
[1]: Joe, Harry, Multivariate models and dependence concepts, Monographs on Statistics and Applied Probability. 73. London: Chapman and Hall. xviii, 399 p. (1997). ZBL0990.62517.
My attempt is giving me a different answer. Using chain rule gives $$d(\varphi(\varphi^{-1}(u)+\varphi^{-1}(v)))=\varphi'(\varphi^{-1}(u)+\varphi^{-1}(v))~d(\varphi^{-1}(u)+\varphi^{-1}(v))=\varphi'(\varphi^{-1}(u)+\varphi^{-1}(v)) ~[\varphi' \circ \varphi^{-1}(u))^{-1} + (\varphi' \circ \varphi^{-1}(v))^{-1}] \, du \, dv$$
Perhaps someone can correct me.