Derivatives and definition

Question

I’m currently doing a course in Mathematical Analysis at University level. I ask myself a simple question; when you’re finding the derivative of a function, you’re essentially finding the rate at which the output is changing (dy) over the rate at which the input is changing (dx). However, whenever my lecturer says the graph is not differentiable at a certain point (eg |x| not differentiable at the origin), they say that there’s isn’t a unique tangent at the origin for that graph. Could someone help me understand what exactly is meant by unique tangent and derivative at a point?

From what I gather: If a function is differentiable at a point, there exists a unique tangent at that point of the graph.

Answer 1

We say that a straight line $l$ in $\Bbb R^2$ is tangent to the graph of $f$ at $a\in D_f$ (the domain of $f$) if, $\bigl(a,f(a)\bigr)\in l$ and if, at least near $a$, $\bigl(a,f(a)\bigr)$ is the only point which belongs both to $l$ and to the graph of $f$.

If $f$ is differentiable at $a$, there is one and only one such tangent line: the graph of the function $t\mapsto f(a)+f'(a)(t-a)$ (which is a straight line).

But if $f(x)=|x|$ then there are infinitely many tangent lines to the graph of $f$ at $(0,0)$. One of them is $y=0$. Another one is $y=x/2$.

Answer 2

Saying that $f$ admits a derivative $a$ at a point $x$ is saying that the ratio $\dfrac{f(x + \varepsilon) - f(x)}{\varepsilon}$ converges to $a$ when $\varepsilon$ goes to zero. Multiply both terms by $\varepsilon$ and add $f(x)$ : you are saying that for little $\varepsilon$, $f(x+\varepsilon) = f(x) + a \varepsilon + \text{ something going to zero}$ : thus the graph of $f$ looks like the line with equation $\varepsilon \mapsto f(x) + a \varepsilon$. In fact, from the very definition of the limit, it is the only line which looks like the graph of $f$ around $x$!

Thus, if your graph looks like two different lines at a point, it cannot have a derivative this point.

That is why $|x|$ has no derivative at $0$ : it looks like $x \to -x$ and $x \to x$ around $0$, which are two different lines.