What does it mean for partial derivative to be continuous and how does that imply differentiability? In order for function to be differentiable at some point, it should be well approximated at that point. I understand that partial derivatives must exist, and that function needs to be continuous, but why partial derivatives must be continuous? 
 A: The statement is not true. 
If a function has continuous partial derivatives on an open set U, then it is differentiable on U. But a differentiable function need not have continuous partial derivatives.
A standard example is the function $f(x)=x^2\sin(\frac1x)$ which is differentiable but its partial derivative with respect to x $f'(x)=2x\sin(\frac1x)-\cos(\frac1x)$ is not continuous.
For the other direction let $f:\mathbb{R}^n\to \mathbb{R}$ have continuous partial derivatives on a neighbourhood $U$ of $p$. Define a linear function
$$F_p(x_1,\dots x_n)=\sum_{i=1}^nx_i\partial f(p)$$
For $x\in U$
$$\begin{align*} f(x)-f(p) &= f(x_1,\dots , x_n)-f(p_1,\dots,p_n) \\
&=f(x_1,\dots , x_n)-f(p_1,x_2,\dots x_n) \\
&+ f(p_1,x_2,\dots, x_n)-f(p_1,p_2,x_3,\dots , x_n) \\
&\qquad \vdots \\
&+ f(p_1,\dots ,p_{n-1},x_n)-f(p_1,\dots ,p_n)\end{align*}$$
Applying the Mean Value Theorem on every line gives:
$$\begin{align*}f(x)-f(p) &=\partial_1f(c_1,x_2,\dots ,x_n)(x_1-p_1) \\
&+\partial_2f(p_1,c_2,x_3,\dots ,x_n)(x_2-p_2) \\
&\qquad \vdots \\
&+\partial_nf(p_1,\dots ,p_{n-1},c_n)(x_n-p_n)\end{align*}$$
With $c_i$ between $x_i$ and $p_i$. Therefore:
$$\frac{f(x)-f(p)-F_p(x-p)}{\|x-p\|}=\sum \underbrace{\frac{x_i-p_i}{\|x-p\|}}_{\text{bounded}}\underbrace{(\partial_if(p_1,\dots ,c_i,\dots x_n)-\partial_if(p))}_{\to 0}$$
A: A map $f:\textbf{R}^n\rightarrow \textbf{R}^m$ which has a differential at $a$ is said to be differentiable at $a$. 
$f$ has a differential, $df(a)$ at $a$ if 
$$
|f(x) - f(a) + df(a)(x-a)|=o(|x-a|)
$$
Where
$$
lim_{x\rightarrow a} \frac{o(|x-a|)}{|x-a|}=0
$$
This is exactly what is shown above, just replace $df(a)$ with $F_a$.
