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The derivative

The derivative at 𝑎𝑎 is the slope of the tangent, written as a limit of slopes of secants. If 𝑓𝑓 is defined near 𝑎𝑎, I set

𝑓′(𝑎)=lim⁡h→0𝑓(𝑎+h)−𝑓(𝑎)h{𝑓}^{\prime }(𝑎)=\lim _{h\to 0}\frac{𝑓(𝑎+h)-𝑓(𝑎)}{h}
(2.1)

when that limit exists. The increment hh is allowed to approach from both sides; if the two one-sided limits disagree, there is no derivative.

The difference quotient is a function of hh, for h≠0h\neq 0 small enough that 𝑎+h𝑎+h stays in the domain. Differentiability at 𝑎𝑎 is a statement about that function as h→0h\to 0, not a statement about 𝑓𝑓 at points far from 𝑎𝑎.

A computation that is actually a computation

For 𝑓(𝑥)=𝑥2𝑓(𝑥)={𝑥}^{2} the algebra is honest rather than mysterious:

𝑓(𝑎+h)−𝑓(𝑎)h=(𝑎+h)2−𝑎2h=2𝑎h+h2h=2𝑎+h.\frac{𝑓(𝑎+h)-𝑓(𝑎)}{h}=\frac{{(𝑎+h)}^{2}-{𝑎}^{2}}{h}=\frac{2𝑎h+{h}^{2}}{h}=2𝑎+h.
(2.2)

Send h→0h\to 0 and 𝑓′(𝑎)=2𝑎{𝑓}^{\prime }(𝑎)=2𝑎, which is (2.1) for this 𝑓𝑓. The hh in the denominator cancelled against a genuine factor of hh in the numerator, not against a wish. The graph of that parabola is Figure 1.1.

A family of secants.
Figure 2.1 Secants approaching a tangent

The same pattern works for 𝑓(𝑥)=𝑥3𝑓(𝑥)={𝑥}^{3}:

(𝑎+h)3−𝑎3h=3𝑎2h+3𝑎h2+h3h=3𝑎2+3𝑎h+h2,\frac{{(𝑎+h)}^{3}-{𝑎}^{3}}{h}=\frac{3{𝑎}^{2}h+3𝑎{h}^{2}+{h}^{3}}{h}=3{𝑎}^{2}+3𝑎h+{h}^{2},

so 𝑓′(𝑎)=3𝑎2{𝑓}^{\prime }(𝑎)=3{𝑎}^{2}. I am not yet claiming a general power rule; I am noticing that a polynomial identity supplies the factor of hh that the definition demands.

Linear approximation

Another way to say the same thing: 𝑓𝑓 is differentiable at 𝑎𝑎 when there is a number 𝑚𝑚 such that

𝑓(𝑎+h)=𝑓(𝑎)+𝑚h+𝑟(h),with𝑟(h)h→0 as h→0.𝑓(𝑎+h)=𝑓(𝑎)+𝑚h+𝑟(h),\quad \text{with}\quad \frac{𝑟(h)}{h}\to 0\, \text{as}\, h\to 0.
(2.3)

Then 𝑚𝑚 is necessarily 𝑓′(𝑎){𝑓}^{\prime }(𝑎), and 𝑓(𝑎)+𝑓′(𝑎)h𝑓(𝑎)+{𝑓}^{\prime }(𝑎)h is the best linear guess for 𝑓𝑓 near 𝑎𝑎. The error 𝑟(h)𝑟(h) is smaller than linear. If I rewrite (2.2) this way, 𝑟(h)=h2𝑟(h)={h}^{2} for 𝑓(𝑥)=𝑥2𝑓(𝑥)={𝑥}^{2}, and 𝑟(h)h=h→0\frac{𝑟(h)}{h}=h\to 0.

Proposition 2.1
If 𝑓𝑓 is differentiable at 𝑎𝑎, then 𝑓𝑓 is continuous at 𝑎𝑎.
Proof
From (2.3), 𝑓(𝑎+h)−𝑓(𝑎)=𝑚h+𝑟(h)𝑓(𝑎+h)-𝑓(𝑎)=𝑚h+𝑟(h). Divide by 11, or rewrite as h(𝑚+𝑟(h)h)h\left(𝑚+\frac{𝑟(h)}{h}\right). As h→0h\to 0, the term in parentheses stays bounded (it tends to 𝑚𝑚) and is multiplied by hh, so the difference tends to 00.

A function that fails

Absolute value at 00 is the standard refusal. The difference quotient is

∣h∣−∣0∣h=∣h∣h={1if h>0−1if h<0\frac{|h|-|0|}{h}=\frac{|h|}{h}=\begin{cases}1\quad & \text{if}\, h>0 \\ -1\quad & \text{if}\, h<0\end{cases}

The right-hand limit is 11, the left-hand limit is −1-1. No single 𝑚𝑚 works, so 𝑥↦∣𝑥∣𝑥\mapsto |𝑥| is not differentiable at 00, even though it is continuous there. Continuity of 𝑓𝑓 is necessary for 𝑓′{𝑓}^{\prime } to exist, by Proposition 2.1, and it is not enough.

Example 2.1
∣𝑥∣|𝑥| is continuous at 00 and not differentiable there.
Remark
Continuity is necessary for 𝑓′(𝑎){𝑓}^{\prime }(𝑎) to exist, and it is not enough.
Exercise 2.1
Prove that ∣𝑥∣|𝑥| fails to be differentiable at 00.
Solution to Exercise 2.1
The difference quotient is +1+1 for h>0h>0 and −1-1 for h<0h<0, so the two one-sided limits differ. A fuller treatment of one-sided derivatives is in Rudin (1976).