Assume $\varphi : I \rightarrow [0, \infty]$ is continuous and strictly decreasing, also inverse here is assumed as pseudo-inverse.
Archimedean copula is defined as $$ C(u,v) = \varphi^{-1}(\varphi(u) + \varphi(v)) = \varphi(\varphi^{-1}(u) + \varphi^{-1}(v))$$ here both definitions hold, depending which is more convenient. Also, $$ c_{u}(v) := \frac{\partial C}{\partial u}(u,v) $$
I'm trying to show that: $$ c_{u}(t)^{-1} = \varphi^{-1}(\varphi(w) - \varphi(u)),$$
here $w = \varphi ' ^{-1}(\varphi'(u)/t)$.
My solution: If I understand correctly,
$$ c_u(v) = (\varphi^{-1})'(\varphi(u) + \varphi(v)) \cdot \varphi'(u)$$ hence
$$ t = c_u( c_u^{-1}(t)) = (\varphi^{-1})'(\varphi(u) + \varphi(c_u^{-1}(t))) \cdot \varphi'(u)$$ $$ \frac{t}{ \varphi'(u)} = (\varphi^{-1})'(\varphi(u) + \varphi(c_u^{-1}(t))) $$ $$ (\varphi^{-1})'^{-1}\left(\frac{t}{ \varphi'(u)}\right) = \varphi(u) + \varphi(c_u^{-1}(t))$$ $$ \varphi(c_u^{-1}(t)) = (\varphi^{-1})'^{-1}\left(\frac{t}{ \varphi'(u)}\right) - \varphi(u) $$ $$ c_u^{-1}(t) = \varphi^{-1}\left( (\varphi^{-1})'^{-1}\left(\frac{t}{ \varphi'(u)}\right) - \varphi(u)\right) $$
Which seems similar to the wanted result, but not exactly the same. I'm not confident in nested derivation, plus not sure how does differentiation and inversion stack, i.e., in the case of $(\varphi^{-1})'^{-1}$ and etc.
What am I missing? Any hints or tips would be appreciated! Also, in order to avoid the XY problem, I may edit the question with additional background on the exercise, but I currently think I'm just missing some understanding in the technical derivation/inversion part.
By the inverse function theorem,
$$(\varphi^{-1})'(x)=\frac{1}{\varphi'(\varphi^{-1}(x))}$$
This means
$$c_u(v) = \frac{ \varphi'(u)}{\varphi'(\varphi^{-1}[\varphi(u) + \varphi(v)]) }$$
So, proceeding as you did:
$$t=c_u(c_u^{-1}(t))=\frac{ \varphi'(u)}{\varphi'(\varphi^{-1}[\varphi(u) + \varphi(c_u^{-1}(t))]) }$$
Hence
$$\varphi'(\varphi^{-1}[\varphi(u) + \varphi(c_u^{-1}(t))])=\frac{\varphi'(u)}{t}$$
Then applying $\varphi^{\prime-1}$ to both sides:
$$\varphi^{-1}[\varphi(u) + \varphi(c_u^{-1}(t))]=\varphi^{\prime-1}\left(\frac{\varphi'(u)}{t}\right)$$
So, using the definition of $w$ and applying $\varphi$ to both sides:
$$\varphi(u) + \varphi(c_u^{-1}(t))=\varphi(w)$$
from which the result follows.