Can you also check this out for me;

$\displaystyle y=e^a$

where $\displaystyle a=-x^2/2$

1...$\displaystyle dy/dx = dy/du \cdot du/dx$

2...where $\displaystyle y = e ^-^u$ and $\displaystyle u= - x^2/2$

3.... $\displaystyle dy/du=e ^-^u$ and $\displaystyle du/dx= -x$ <<<< can you explain how step 2 for "u" this turns into -x

4.... therefore; $\displaystyle dy/du \cdot du/dx = -x e ^ - x^2/2$