Directional Derivative and Level Sets Orthogonality My book states without proof that the directional derivative at any point is orthogonal to the tangent to the level set at the same point. 
I don't even know where to get started.
All I contribute is that :
Assume $f : R^n \rightarrow R$ (I can make this assumption as per question)


*

*$D_v f(a_1,a_2,\cdots,a_n) = ||\nabla{f(x,y)}||_{a_1,a_2,\cdots,a_n} \cdot \dfrac{v}{||v||}$

*I need to show that 2 vectors are orthogonal and thus, I feel that there is a point where I'd need to show that the inner product of the above directional derivative vector with the tangent vector is 0. (But what is the tangent vector?)
 A: A differentiable function $f:\ {\mathbb R}^n\to {\mathbb R}$ has at each point ${\bf p}$ of its domain a gradient
$$\nabla f({\bf p})=\Bigl({\partial f\over\partial x_1},{\partial f\over\partial x_2},\ldots,{\partial f\over\partial x_n} \Bigr)_{\bf p}\ .$$
On the other hand, the directional derivative of $f$ at ${\bf p}$ in direction ${\bf v}$ is given by
$$D_{\bf v}f({\bf p}):=\lim_{t\to0}{f({\bf p}+ t{\bf v})-f({\bf p})\over t} =\nabla f({\bf p})\cdot{\bf v}\ .$$
Now the level surface: If $\nabla f({\bf p})\ne{\bf 0}$ then the level set of $f$ through the point ${\bf p}$ is locally a smooth surface $S$. Consider a curve
$\gamma:\ t\mapsto {\bf x}(t)$ drawn on $S$ with ${\bf x}(0)={\bf p}$. Then the function
$$\phi(t):=f\bigl({\bf x}(t)\bigr)$$
is constant, namely $\equiv f({\bf p})$. It follows that $\phi'(t)\equiv0$. In particular we have by the chain rule 
$$0=\phi'(0)=\nabla f\bigl({\bf x}(0)\bigr)\cdot {\bf x}'(0)=\nabla f({\bf p})\cdot {\bf x}'(0)\ ,$$
which says that $\nabla f({\bf p})$ is orthogonal to the tangent vector ${\bf x}'(0)$ to $\gamma$ at ${\bf p}$. Since this is true for any level curve $\gamma$ through ${\bf p}$ it follows that the vector $\nabla f({\bf p})$ is orthogonal to the level surface $S$.
A: The answer is a bit involved. You can read it on page 335 of this link. The idea is to parametrize level curves and to differentiate implicitly the equation $f(x,y)=c$.
