For example the function $f(x)=|x|$, the graphic will be something like the letter "V". The function is continuous in $x=0$. However, the slope is different for $x < 0$ than for $x>0$, just like in quadratic functions, so I don’t see how that explains it. Can’t I draw a tangent line to the graph in $x=0$ which coincides with the $x$ axis? If i am able to do that, why isn’t the derivative (in $x=0$) equal to zero as in quadratic functions?
[Math] Why is there no derivative in an absolute value function
calculusderivatives
Related Solutions
No, the derivative $f'(a)$ is the exact slope of the line tangent to $(a, f(a))$. Note that by definition
$$f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$$
What the article is saying is that for very small $|h| > 0$, the slope of the secant line passing through $(a, f(a))$ and $(a + h, f(a + h))$ is a good approximation of $f'(a)$.
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.
Best Answer
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?