Let $f: \mathbb{R}^{l} \to \mathbb{R}^{k}$ be continuously differentiable, and let $S=f^{-1}(0)$. Assume that, for some $y \in S$, $f'(y)$ has rank $k$. Show that $T(y,S)=\text{ker} f'(x)$, where $T(y,S)$ is the tangent cone of $S$ at $y$.


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