Why is the differential being onto equivalent to it not being zero?

[yifli]

I have difficulty understanding the following Theorem

If U is open in ℝ^2, F: U \rightarrow ℝ is a differentiable function with Lipschitz derivative, and X_c=\{x\in U|F(x)=c\}, then X_c is a smooth curve if [\operatorname{D}F(\textbf{a})] is onto for \textbf{a}\in X_c; i.e., if \big[ \operatorname{D}F\bigl( \begin{smallmatrix}a \\ b\end{smallmatrix}\bigr)\big]≠0 \mbox{ for all } \textbf{a}=\bigl( \begin{smallmatrix}a \\ b \end{smallmatrix}\bigr)\in X_c

I don't understand why the differential of F at a being onto is equivalent to saying the differential is not zero. Can someone explain? Thanks

 

[HallsofIvy]

Think about what happens for simple functions like f(x)= x^2 where f'(x)= 0.

 

[yifli]

Think about what happens for simple functions like f(x)= x^2 where f'(x)= 0.

the differential of f(x)=x^2 is 2x, so f'(x)=0 means x=0.
Now what?