A definition of the derivative is that it is the slope of the tangent line. For example, $x^3$ has a quadratic derivative. How could the slope of the tangent line be non-linear?
[Math] Why can a derivative be non-linear
calculusderivatives
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The second derivative tells you something about how the graph curves on an interval.
If the second derivative is always positive on an interval $(a,b)$ then any chord connecting two points of the graph on that interval will lie above the graph. If the second derivative is always negative on the interval $(a,b)$ then any chord connecting two points of the graph on that interval will lie below the graph.
In the graph below of $y=x(x-1)(x+1)$ the graph has a negative second derivative on the interval $(-\infty,0)$ and a positive second derivative on the interval $(0,\infty)$ so it is concave down and concave up, respectively on the two intervals.
Another way of expressing the same idea is that if a continuous second differentiable function has a positive second derivative at point $(x_0,y_0)$ then on some neighborhood of $(x_0,y_0)$ the tangent line at $(x_0,y_0)$ lies below the graph (except at the point of tangency). If the second derivative is negative at the point of tangency the tangent line lies above the graph on some neighborhood of the point of tangency (except at the point of tangency).
You can think this geometrically. The derivative of a one variable function is the slope of the tangent line. The slope, which is defined as a limit, will exist and will be unique if there is only one tangent line. Now in case of $f(x)=|x|$, there is no one unique tangent at $0$. I refer to you to the following graph : 
Now ask yourself which "tangent" you want to consider?
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Edited
OK, so I'll try to make this post more self contained.
To find an equation of the tangent line to any function $y = f(x)$ at the point $A(x_0, y_0)$, iff $f(x)$ is differentiable at that point, one needs to calculate its algebraic form as $$ y = y_0 + f'(x_0)(x - x_0) $$ so, as you can see, even though you calculate $f'(x)$ to find tangent line slope, it's evaluated at $x_0$, which makes it a number, but that number will change when you move from point to point.
Now, let's consider that example $$ y = x^3 \implies y' = 3x^2 \implies y = x_0^3 + 3x_0^2(x - x_0) = 3x_0^2 x - 2x_0^3 $$ Simple animation of the tangent line when point it was calculated at is moving provided below.