Can directional derivatives be written as linear combination of partial derivatives even if f is not differentiable? Given $f:\mathbb{R}^n\to\mathbb{R}$ and $\{v_1,...,v_n\}$ linearly independent vectors such that $\frac{\partial f}{\partial v_i}$ exists.
I know if f is differentiable then $\frac{\partial f}{\partial v_i}$=$\nabla f\cdot v$ so the directional derivative of f can be expressed as a linear combination of the partial derivatives.
If f is not differentiable: 
1) $\nabla f$ exists ?
2) If $\nabla f$ exists ,can we express the directional derivatives as a linear combination of the partial derivatives?
 A: There are many examples of functions — even discontinuous functions — that have all partial derivatives but for which that linearity formula fails. You can find many of them littered around in questions on this site.  But here are a few. Set $f(0,0) = 0$ and for $(x,y)\ne (0,0)$ take
\begin{align*}
f(x,y) &= \frac{xy}{x^2+y^2} \\
f(x,y) &= \frac{xy^2}{x^2+y^4} \\
f(x,y) &= \frac{xy^2}{x^2+y^2}
\end{align*}
Find the gradient vector (hint: they'll all be $0$) at the origin, and compute the various directional derivatives.
A: Some other examples:
Let  $f(x) = 0 $ on the axes, but $1$ everywhere else. Then the partials along the axes exist at the origin but the function is not continuous at the origin.
Let $A = \{ (r \cos t, r \sin t)  | 0 < t \le 2 \pi, 0 \le r  \le t\}$ and $f=1_A$. Then the directional derivative $df(0,h) = 0$ for all directions $h$, but $f$ is not continuous at the origin.
The point of the last example is a function where the directional derivative
can be written as a linear combination of the partial derivatives but the
function is not continuous (and hence not differentiable) at the point.
