How we simplify equation with derivation

Question

Our teacher write :

$(zz')' +2xz' = 0,\quad 0 < x < \infty \tag {1}$

We can write equation $(1)$ like this

$(z^2)'' + 4xz' =0,\quad 0 < x < \infty \tag{2}$

but I can not understand how we find the equation $(2).$

Answer 1

The key observation here is that $$ (z^2)'' = 2(zz')' $$ which can be derived using either the product rule or the chain rule. Using the chain rule, if we let $f(x) = x^2$, then we're looking to find $(f(z))' = f'(z)z'$, and since $f'(x) = 2x$, this is just $2zz'$. To get the second derivative, just apply the derivative again (but don't simplify): $(f(z))'' = (f'(z)z')' = (2zz')' = 2(zz')'$, where the last equality is true since the derivative is a linear operator.

This means, if we multiply the first equation through by $2$, then we get $$ 2(zz')' + 4xz' = (z^2)' + 4xz' $$ as expected.

Answer 2

Application of the Chain Rule :

$$(z^2)’ = 2zz’$$

Thus:

$$(z^2)’ + 4xz = 0$$

becomes

$$2(zz’)’ + 4xz =0$$

Divide through by 2 to get your first equation.