2nd Order Homo ODE Obtain Basis When One Solution Known [Reduction of Order]

Question

According to my text:

If we know one solution $y_1$ to $y^{"} + p(x)y' + q(x)y = 0$ then a second independent solution $y_2$ can be found if we perform a reduction of order by substituting $y(x) = u(x)y_1(x)$.

$$y' = u'y_1 + uy_1'\\ y^{"}= u^{"}y_1 + 2u'y_1' + uy_1^{"}$$

There is more to this derivation but I am not interested in it. My question is where did the 2 come from in the second order differential equation above?

Answer 1

$$ y' = u'y_1 + uy_1'\\ y^{"}= \frac{d}{dx}(u'y_1+ uy_1')=\underbrace{u^{"}y1 + u'y_1'}_{\frac{d}{dx}(u'y_1)}+\underbrace{u'y_1' + uy_1^{"}}_{\frac{d}{dx}(uy_1')}=u^{"}y_1 + 2u'y_1'+ uy_1^{"} $$