Legendre Differential Equation (rearrange)

Question

Legendre differential equation: $$ (1-x^2) \, y'' - 2x y' + \alpha(1+\alpha)y = 0 \, $$

which it can also be written as:
$$ \frac{\mathrm{d}}{\mathrm{d}x} \left( (1-x^2) \frac{\mathrm{d}y}{\mathrm{d}x} \right) + \alpha(1+\alpha)y = 0 \ $$

how did we reach equation two from equation one. I don't get it please help

Answer 1

Apply the basic rule $\frac{dfg}{dy} = f\frac{dg}{dx}+g\frac{df}{dx}$

So $\frac{d(1-x^2)\frac{dy}{dx}}{dx} = 1-x^2*y'' -2x*y'$

Answer 2

$\frac{d}{dx}(\frac{dy}{dx}) = \frac{d^2y}{dx^2} = y’’$