Given point and normal vector find the derivative of a function

Question

1. State the properties that a $f$ function must have to ensure that its directional derivative $$\begin{matrix}f'\left(\vec A,\check v\right)&=&\nabla f\left(\vec A\right)\cdot\check v&\forall\check v.\end{matrix}$$ 2. Calculate $f'\left(\vec A,(0.6,0.8)\right)$ with $\vec A=(x_0,y_0)$ knowing that $\vec n_0=(2,-3,5)$ is a vector normal to the surface of equation $z=f(x,y)$ at the point $(x_0,y_0,f(x_0,y_0))$.

---

1. I said that $f:A\subseteq\mathbb R^n\to\mathbb R$ must be differentiable at $\vec A\in\color{red}{A^0\;(\text{or }A?)}$. Or I have to say only for the case where $n=2$? What is your opinion?
2. We know that by statement $f$ is differentiable at $\vec A$, so we can apply the previous property. So let $\check v=(v_1,v_2)$. Then $$f'\left(\vec A,\check v\right)=\nabla f\left(\vec A\right)\cdot\check v=\left(f'_x(\vec A),f'_y(\vec A)\right)\cdot(v_1,v_2)=f'_x(\vec A)\cdot v_1+f'_y(\vec A)\cdot v_2=2\cdot0.6-3\cdot0.8=\boxed{-1.2}.$$

Is that OK? Could you help me with the first item, please?

Thank you!

Answer 1

${\bf 0.}$ The notations of your source are absolutely horrible. I won't stick to them.

${\bf 1.}$ I think it is tacitly assumed that your function $f$ is defined in some open region $\Omega\subset{\mathbb R}^2$. For the intended computations one has to assume that $f$ is differentiable at the point ${\bf a}=(x_0,y_0)\in \Omega$. A sufficient condition for this to be the case is that the partial derivatives $f_x$ and $f_y$ exist in a neighborhood of ${\bf a}\in\Omega$, and are continuous at ${\bf a}$.

${\bf 2.}$ The surface $S$ in question is given by the equation $$\bigl(\ F(x,y,z):=\ \bigr)\quad f(x,y)-z=0\ .$$ If ${\bf p}:=(x_0,y_0,z_0)\in S$ then the gradient $$\nabla F(x_0,y_0,z_0)=\bigl(f_x(x_0,y_0), f_y(x_0,y_0),-1)\tag{1}$$ is $\ne{\bf 0}$, hence is orthogonal to the tangent plane $T_{\bf p}S$. Since we are told that the vector ${\bf n}_0=(2,-3,5)$ is orthogonal to $T_{\bf p}S$ as well we may infer that $\nabla F({\bf p})$ and ${\bf n}_0$ are proportional. Comparing the third components we therefore conclude that$$\nabla F(x_0,y_0,z_0)=-{1\over 5} {\bf n}_0=\left(-{2\over5},{3\over5},-1\right)\ .$$ It follows that $$\nabla f({\bf a})=\bigl(f_x(x_0,y_0), f_y(x_0,y_0)\bigr)=\left(-{2\over5},{3\over5}\right)\ .\tag{2}$$ Note that in $(1)$ the $\nabla$ referred to $(x,y,z)$, while in $(2)$ it refers only to $(x,y)$!

In the case at hand we are not given the coordinates $(x_0, y_0)$ of the point ${\bf a}\in {\mathbb R}^2$, but we now know that $\nabla f({\bf a})$ is given by $(2)$. This allows to conclude that $$df({\bf a}).(0.6,0.8)=\nabla f({\bf a})\cdot(0.6,0.8)=-{2\over5}0.6+{3\over5}0.8=0.24\ .$$