Suppose $f: \mathbb{R}^n \to \mathbb{R}$ is differentiable at $x$. Let $d_xf$ denote the derivative of $f$ at $x$. Let $L$ be the level set through $x$, $L = \{y \in \mathbb{R}^n: f(y) = f(x)\}$. Suppose $v$ is a tangent vector at $x$ that is tangent to the level set $L$. Then the claim is that $d_xf(v) = 0$. Why is this true? Is there a rigorous justification of this?
I have seen various answers, like Why is the gradient normal? and Why gradient vector is perpendicular to the plane, but I couldn't really find a rigorous justification of that particular fact, that $d_xf(v) = 0$. I can see the intuition but I'd like a proof if possible.
Also, what does it mean precisely when we say "Suppose $v$ is a tangent vector at $x$ that is tangent to the level set $L$"? Can all these facts and notions be defined and proved in the usual multivariable context of Euclidean space $\mathbb{R}^n$ (like in a normal or advanced Calc III course) or do we need an excursion into differential geometry or something? I'd just like to know because some of the multivariable calculus texts/resources I've seen, as well as some answers on this site, mostly seem to gloss over the details and just roughly justify it by appealing to geometric intuition, which I think is useful but I would also like a proof.
Best Answer
If you're granting the fact (given by the implicit function theorem) that the level set actually has a tangent plane at $x$, then any tangent vector is the velocity vector of some curve $\gamma(t)$ contained in the level set. We may assume that $\gamma(0)=x$ and $\gamma'(0)=v$. Then $f(\gamma(t)) = \text{constant}$ (by definition of level set), and so, by the chain rule, $$0=d_xf(\gamma'(0)) = d_xf(v),$$ as you desired.