Prove that $\frac{d}{dx}(xJ_\alpha(x)J_{\alpha+1}(x))=x(J_\alpha^2(x)-J_{\alpha+1}^2(x))$

Here is an easy method using only the two identities you are given at the start of the question.

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Notice that: $$\color{red}{x^{\alpha+1}J_{\alpha+1}(x)}\cdot \color{blue}{x^{-\alpha} J_\alpha(x)}=xJ_{\alpha}(x)J_{\alpha+1}(x)$$ Therefore, we can apply the product rule with $u=x^{\alpha+1}J_{\alpha+1}(x)$ and $v=x^{-\alpha}J_{\alpha}(x)$: $$\begin{align}(uv)'&=u'v+uv'\\&=(x^{\alpha+1}J_{\alpha+1}(x))'\cdot x^{-\alpha}J_{\alpha}(x)+x^{\alpha+1}J_{\alpha+1}(x)\cdot (x^{-\alpha}J_{\alpha}(x))' \end{align}$$ Substituting (i) and (ii) and simplifying gives the required identity: $$x^{\alpha+1}J_{\alpha}(x)\cdot x^{-\alpha}J_{\alpha}(x)+x^{\alpha+1}J_{\alpha+1}(x)\cdot -x^{-\alpha}J_{\alpha+1}(x)=x(J_{\alpha}^2(x)-J_{\alpha+1}^2(x))$$