proving that $\frac{d}{dx}\phi_j(x)=\frac{4}{l}\sum_{m=0 \\ (j+m) odd}^{j-1}\frac{3j-2m}{\xi_m}\phi_m(x)+\eta^l_j(x)$

Question

Given that $$\frac{d}{dx}T^l_k(x)=\frac{1}{l}\sum_{m=0 \\(k+m)odd}^{k-1}\frac{4k} {\xi_m}T^l_m(x)$$ AND $$\phi_j(x)=x(l-x)T^l_j(x)$$ Where $$ \xi_m = \left\{ \begin{array}{ll} 2 & m = 0 \\ 1 & otherwise \end{array} \right. $$ Prove that $$\frac{d}{dx}\phi_j(x)=\frac{4}{l}\sum_{m=0 \\ (j+m) odd}^{j-1}\frac{3j-2m}{\xi_m}\phi_m(x)+\eta^l_j(x)$$ where $$\eta^l_j(x) = \left\{ \begin{array}{ll} l-2x &j \ \ \ even \\ -l & j \ \ \ odd \end{array} \right.$$ What i have got so far is $$\frac{d}{dx}\phi_j(x)=(l-2x)T^l_k(x)+\frac{4}{l}\sum_{m=0 \\(j+m) odd}^{j-1}\frac{j}{\xi_m}\phi_m(x)$$ Edited $$T^l_k(x) =\cos(k \cos^{-1}(2x/l -1 ))$$ How to get the required result?