Given the following Gevrey-class function $\Phi:\mathbb{R} \rightarrow \mathbb{R}$
$$\Phi_{s,T}(t) = \begin{cases} \begin{align} 0 \quad & t \le 0 \\ 1 \quad & t \ge T \\ \frac{\int\limits_{0}^t\Lambda_{s,T}(\tau) \text{d}\tau}{\int\limits_{0}^T\Lambda_{s,T}(\tau) \text{d}\tau} \quad & t \in (0,T), \end{align} \end{cases} $$ where $$\Lambda_{s,T}(t) = e^{-\frac{1}{[(1-t/T)t/T]^s}}, \quad s \in \mathbb{N^+}, \: T\in\mathbb{R}. $$
I know there must be a recurrence for the $(k+1)^{\text{st}}$-derivative of $\Phi_{s,T}$:
$$\Phi_{s,T}^{(k+1)}(t) = f\left(\Phi_{s,T}^{(k)}(t), \Phi_{s,T}^{(k-1)}(t)\right), \quad t \in (0,T).$$ Does anybody know what it is and how it can be proven inductively?
Lemma. The $(k+1)^{\text{st}}$ derivative of $\Phi_{s,T}$ is recursivly defined as follows: $$ \begin{align} \Phi_{s,T}^{(k+1)}(t) & = \frac{\Lambda_{s,T}^{(k)}(t)}{\int\limits_0^T\Lambda_{s,T}(\tau)\text{d}\tau}, \quad k=0,...,n_t\\ \Lambda_{s,T}^{(k+1)}(t) & = \sum\limits_{j=0}^{k} \left( \begin{array}{c} k \\ j \end{array} \right)\Lambda_{s,T}^{(k-j)}(t)\lambda^{(j+1)}(t) \\ \Lambda_{s,T}^{(0)}(t) = \Lambda_{s,T}(t) & = e^{-p(t)^{-s}}\\ \lambda^{(j+1)}(t) & = p^{-1}(t)\left(\lambda^{(j)}(t)p'(t)(-s-j)-\lambda^{(j-1)}(t)p''(t)j\left(s+\frac{j-1}{2}\right)\right), \quad j=1,...,k \\ \lambda^{(0)}(t) = \lambda(t) & = -p(t)^{-s} \\ \lambda^{(1)}(t) = \lambda'(t) & = sp(t)^{-s-1}p'(t) \\ p(t) &= \left(1-\frac{t}{T}\right)\frac{t}{T}. \end{align} $$ Proof. By induction. $$ \Lambda_{s,T} \left\{ \begin{align} (k=0)\qquad & \Lambda_{s,T}^{(1)}(t) = \left( \begin{array}{c} 0 \\ 0 \end{array} \right) \Lambda_{s,T}^{(0)}(t)\lambda^{(1)}(t) = e^{-p(t)^{-s}}sp(t)^{-s-1}p'(t) \\ (k \implies k+1) \quad & \left(\Lambda_{s,T}^{(k)}(t)\right)' = \left( \sum\limits_{j=0}^{k-1} \left( \begin{array}{c} k-1 \\ j \end{array} \right)\Lambda_{s,T}^{(k-1-j)}(t)\lambda^{(j+1)}(t) \right)' \\ & = \sum\limits_{j=0}^{k-1} \left( \begin{array}{c} k-1 \\ j \end{array} \right) \left( \Lambda_{s,T}^{(k-1-j)}(t)\lambda^{(j+1)}(t) \right)' \\ & = \sum\limits_{j=0}^{k-1} \left( \begin{array}{c} k-1 \\ j \end{array} \right) \left( \underbrace{ \Lambda_{s,T}^{(k-j)}(t)\lambda^{(j+1)}(t) }_{(i)} + \underbrace{ \Lambda_{s,T}^{(k-1-j)}(t)\lambda^{(j+2)}(t) }_{(ii)} \right) \end{align} \right. $$ Note that (ii) of the actual and (i) of the successive sum index match such that the overall sum is given as $$ \begin{align} \left(\Lambda_{s,T}^{(k)}(t)\right)' &= \Lambda_{s,T}^{(k-j)}(t)\lambda^{(j+1)}(t) + \left(1 + \left(\begin{array}{c}k-1\\1\end{array}\right)\right)\Lambda_{s,T}^{(k-j-1)}(t)\lambda^{(j+2)}(t) \\ & + \left( \left( \begin{array}{c} k-1 \\ 1 \end{array} \right) + \left( \begin{array}{c} k-1 \\ 2 \end{array} \right) \right) \Lambda_{s,T}^{(k-j-2)}(t)\lambda^{(j+3)}(t) + ... + \Lambda_{s,T}^{(0)}(t)\lambda^{(k+1)}(t), \end{align} $$ where the coefficients evaluate to $$ \begin{align} \left( \begin{array}{c} k-1 \\ j \end{array} \right) + \left( \begin{array}{c} k-1 \\ j+1 \end{array} \right) & = \frac{(k-1)!}{j!(k-1-j)!} + \frac{(k-1)!}{(j+1)!(k-2-j)!} \\ & = \frac{(k-1)!((j+1)+(k-1-j))}{j!(k-2-j)!(j+1)(k-1-j)} \\ & = \frac{k!}{(j+1)!(k-1-j)!} \\ & = \left( \begin{array}{c} k \\ j+1 \end{array} \right). \end{align} $$ This proves the hypothesis $$ \left(\Lambda_{s,T}^{(k)}(t)\right)' = \sum\limits_{j=0}^{k} \left( \begin{array}{c} k \\ j \end{array} \right)\Lambda_{s,T}^{(k-j)}(t)\lambda^{(j+1)}(t) = \Lambda_{s,T}^{(k+1)}(t). $$ Similarly, we have for $\lambda$ that (arguments omitted) $$ \lambda \left\{ \begin{align} (j=1) \quad & \lambda^{(2)} = p^{-1} \left( \lambda^{(1)}p'(-s-1)-s\lambda^{(0)}p'' \right) \\ (j\implies j+1) \quad & \left(\lambda^{(j)}\right)' = \left( p^{-1} \left( \lambda^{(j-1)}p'(-s-j+1) - \lambda^{(j-2)}p''\sum\limits_{i=0}^{j-2}(s+i) \right)\right)' \\ & = -p^{-2}p' \left( \lambda^{(j-1)}p'(-s-j+1) - \lambda^{(j-2)}p''\sum\limits_{i=0}^{j-2}(s+i) \right) \\ & + p^{-1} \left( (-s-j+1)\left(\lambda^{(j-1)}p'\right)' - \sum\limits_{i=0}^{j-2}(s+i) \left(\lambda^{(j-2)}p''\right)' \right) \\ & = p^{-1}\left(p'\lambda^{(j)}(-1+(-s-j+1)) \\ - \lambda^{(j-1)}p''\left( \sum\limits_{i=0}^{j-2}(s+i)+(s+j-1) \right) \right) \\ & = p^{-1} \left( \lambda^{(j)}p'(-s-j) - \lambda^{(j-1)}p''\sum\limits_{i=0}^{j-1}(s+i) \right) \\ & = \lambda^{(j+1)}. \end{align} \right. $$ $\Box$