Bessel function equation with derivation: $\frac{d}{dx} (xJ_{\alpha}(x)J_{\alpha+1}(x)) = x(J_{\alpha}^2(x)-J_{\alpha +1}^2(x))$ Let $J_{\alpha}(x)$be the Bessel function. Show the equality:
$$\frac{d}{dx} \left( xJ_{\alpha}(x)J_{\alpha +1}(x)\right) = x \left( J_{\alpha}^2(x) -J_{\alpha +1}^2(x)\right)$$
How to start with it?
 A: $$\frac{\text{d}}{\text{d} x}\left(x\text{J}_{\alpha}(x)\text{J}_{\alpha+1}(x)\right)=$$

Use the product rule, $\frac{\text{d}}{\text{d}x}(uv)=v\frac{\text{d}u}{\text{d}x}+u\frac{\text{d}v}{\text{d}x}$, where $u=x$ and $v=\text{J}_{\alpha}(x)$:

$$\text{J}_{\alpha}(x)\text{J}_{1+\alpha}(x)\left[\frac{\text{d}}{\text{d}x}(x)\right]+x\left[\frac{\text{d}}{\text{d}x}\left(\text{J}_{\alpha}(x)\text{J}_{1+\alpha}(x)\right)\right]=$$
$$\text{J}_{\alpha}(x)\text{J}_{1+\alpha}(x)\left[1\right]+x\left[\frac{\text{d}}{\text{d}x}\left(\text{J}_{\alpha}(x)\text{J}_{1+\alpha}(x)\right)\right]=$$
$$\text{J}_{\alpha}(x)\text{J}_{1+\alpha}(x)+x\left[\frac{\text{d}}{\text{d}x}\left(\text{J}_{\alpha}(x)\text{J}_{1+\alpha}(x)\right)\right]=$$

Use the product rule, $\frac{\text{d}}{\text{d}x}(uv)=v\frac{\text{d}u}{\text{d}x}+u\frac{\text{d}v}{\text{d}x}$, where $u=\text{J}_{\alpha}(x)$ and $v=\text{J}_{\alpha+1}(x)$:

$$\text{J}_{\alpha}(x)\text{J}_{1+\alpha}(x)+x\left[\text{J}_{1+\alpha}(x)\cdot\frac{\text{J}_{\alpha-1}-\text{J}_{1+\alpha}(x)}{2}+\text{J}_{\alpha}(x)\left[\frac{\text{d}}{\text{d}x}\left(\text{J}_{1+\alpha}(x)\right)\right]\right]=$$
$$\text{J}_{\alpha}(x)\text{J}_{1+\alpha}(x)+x\left[\text{J}_{1+\alpha}(x)\cdot\frac{\text{J}_{\alpha-1}-\text{J}_{1+\alpha}(x)}{2}+\text{J}_{\alpha}(x)\left[\frac{1}{2}\left(\text{J}_{\alpha}(x)-\text{J}_{2+\alpha}(x)\right)\right]\right]=$$
$$x\left[\text{J}_{\alpha}^2(x)-\text{J}_{1+\alpha}^2(x)\right]$$
