# Circular definition of tangent line and derivative

I'm trying to understand the deep relations between the tangent line to the graph of a function $f$ at a given point $P$, and the derivative of $f$ at the same point.

Indeed, in many books the derivative is often defined as the slope of the "tangent", with only an intuitive definition of what a tangent line is. Then, once the concept of derivative is well assimilated, books define more precisely the tangent as... the line passing through $P = (a, f(a))$ with slope $f'(a)$ ! This seems a bit circular.

So some authors use an other approach, starting by giving a geometric definition of a tangent, and then showing the connection with derivatives. That's what one can read in the paper "What a tangent line is when it isn't a limit", by Irl Bivens. But I am missing something in the proof of Theorem 1, when the author writes:

it suffices to show that $|f(x)-L(x)|\leq (1/2)|L(x)-K(x)|$

where $f$ is a differentiable function, $L$ is the line of equation $L(x) = f'(a)(x-a)+f(a)$ and $K$ an other line, of equation $K(x)=p(x-a)+f(a)$ with $p \neq f'(a)$. So, basically, the goal is to show that $L$ is a better approximation of $f$ than any $K$.

But where does the right-hand side come from ? It looks like, locally, $f(x) = 1/2 (L(x)+K(x))$, but why?

• Perhaps it would be helpful if you included a little more context here, at the very least describe what the symbols $f, L, K$ are are. – Travis May 30 '15 at 14:36

Given that $$|L(x)-K(x)|=|L(x)-f(x)+f(x)-K(x)|\leq|f(x)-L(x)|+|f(x)-K(x)|\tag1,$$ if $$|L(x)-K(x)|\geq2|f(x)-L(x)|\tag2$$ then $$2|f(x)-L(x)|\leq|f(x)-L(x)|+|f(x)-K(x)|,$$ from which you obtain easily that $$|f(x)-L(x)|\leq|f(x)-K(x)|\tag3.$$
• @Greg82: $(1)$ is a simple application of the triangular inequality, $(2)$ is your inequality, $(3)$ is the inequality defining the tangent line, I show that $(1)+(2)\implies(3)$. – enzotib May 30 '15 at 16:08