# Nondegenerate critical point

I don't understand this part from the book of Zeidler , can someone help me to understand it ? If $u_0$ is a non degenerate critical point for some function $f$ then under a smooth change of coordinates, $u_0$ is still a non degenerate critical point.
• but mhy they say: since $\varphi'(u_0)\neq 0$ , this fact follows easily from the chain rule ,please thank you – Vrouvrou May 4 '13 at 21:59
• The chain rule says $(f\circ \phi)'(u_0) = f'(\phi(u_0))\phi'(u_0) = f'(u_0) \phi'(u_0)$. Now ask yourself what it means to be a critical point of $f$ in different coordinates. – Mud May 4 '13 at 22:15
• $u_0$ is a critical point of f if $f'(u_0)=0$, so if it is a critical point to $f$ it still a critical point of $f\circ\varphi$, to see that it is nondegenerate : $(f\circ\varphi)''(u_0)=f''(u_0)\varphi'(u_0)+f'(u_0)\varphi''(u_0)\neq 0$ since $\varphi'(u_o)\neq0$. right ? – Vrouvrou May 5 '13 at 5:49
• one question please: why $\varphi'(u_0)\neq 0$ ? – Vrouvrou May 5 '13 at 6:14