In calculus, if a function is differentiable at a point $c$, then it is continuous at $c$, but that the converse of this statement is not true (for example, $f(x) = |x|$ at the point $0$).
What does that mean?
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$\begingroup$Differentiability of $f(x)$ at a point $c$ inherentily means it can be differentiated at point $c$, i.e.:$$\color{red}{\lim_{\Delta{x}\to0^-}\frac{f(c+\Delta{x})-f(c)}{\Delta{x}}}=\color{blue}{\lim_{\Delta{x}\to0^+}\frac{f(c+\Delta{x})-f(c)}{\Delta{x}}}=\lim_{\Delta{x}\to0}\frac{f(c+\Delta{x})-f(c)}{\Delta{x}}$$
This is because of first principles and means that a point $c$ is differentiable when the left-hand derivative (red) equals to the right-hand derivative (blue) as this is the definition for a limit to exist. For the function you are talking about:
$$|x|=\begin{cases} x, \ \text{if} \ x \geq 0 \\ -x, \ \text{if} \ x<0 \end{cases}$$
The left-hand derivative is $-1$, the right-hand limit is $1$ and thus it is not differentiable at $c=0$ since $-1\neq{1}$. Therefore, while $|x|$ is continuous at all points, it is not necessarily differentiable since the point $c=0$ is not differentiable. I hope this helps!
$\endgroup$ $\begingroup$"Differentiable" means that the graph is smooth, which means the tangent line at that point is nearly the same as the function.
The graph of $f(x) = |x|$ has a sharp point at $x=0$. If you try to draw a tangent line to the curve at that point, no matter what line you choose, the gap between the line and the curve is sort of big.
But if you do the same thing for the function $g(x) = x^2$ at $x=0$, you can find a line (the tangent line) which is such a good approximation to the curve that, when you're close to zero, you can't tell the difference between the line and the curve.
If you're building a train track, it's good to make it continuous, but it's even better if it's differentiable, because then there will be no sharp corners.
(Note that I used the word "smooth" above. I'm not using this is the technical sense.)
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