I was reading the book of optimizaion by Polyak and I found this definition:
A scalar function $f(x)$ of an $n$-dimentional argument $x$ ($f:\mathbb{R}^n\rightarrow\mathbb{R}$) is said to be differentiable at a pint $x$ if we can find a vector $a\in\mathbb{R}^n$ such that for all $y\in\mathbb{R}^n$,$$f(x+y)=f(x)+a\cdot y+o(y)$$ The vector $a$ is called the derivative or the gradient of $f(x)$ at a point $x$ and is written $f'(x)$ or $\nabla f(x)$.
$o(y)$ is a function such that $$\lim\limits_{||y||\rightarrow 0}\dfrac{||o(y)||}{||y||}=0.$$
My question is if this definition is equivalen to said that $\nabla f(x)=\begin{pmatrix}\dfrac{\partial f}{\partial x_1},\cdots,\dfrac{\partial f}{\partial x_n}\end{pmatrix}$, because this is easier, or it is just a consequence when the derivative exists? (with this definition), and in this case what would be a fuction where its partial derivates exist but it isn't differentiable?
The vector $ a $ in your definition is exactly $ \nabla f(x) $ where the " $(x)$ " part stress the dependance on the point. So, if $f$ is differentiable at $x$, it has all partial drivatives and happens that $\nabla f(x) = (\frac{\partial f}{\partial x_1},...,\frac{\partial f}{\partial x_n})$ .
A function that has partial derivatives but is not differentiable is $f(x) = \frac{x^2y}{x^4+y^2}$ for $x \neq 0$ and $f(0)=0$ if you take the origin as base point