Differentiability

A function is differentiable at a point if its graph has a tangent line there. Algebraically, this means the limit defining the derivative exists. The derivative measures the instantaneous rate of change , slope of the tangent, velocity from displacement, marginal cost from cost.

Definition

The function $f$ is differentiable at $a$ if

$f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$

exists (as a finite limit). The value of the limit is the derivative of $f$ at $a$.

Equivalent form: $f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}$.

One-sided derivatives

The left derivative at $a$:

$f'_-(a) = \lim_{h \to 0^-} \frac{f(a + h) - f(a)}{h}.$

The right derivative:

$f'_+(a) = \lim_{h \to 0^+} \frac{f(a + h) - f(a)}{h}.$

$f$ is differentiable at $a$ iff both one-sided derivatives exist and are equal.

Differentiability implies continuity

Theorem. If $f$ is differentiable at $a$, then $f$ is continuous at $a$.

Proof. $\lim_{x \to a}[f(x) - f(a)] = \lim_{x \to a}\left[\dfrac{f(x) - f(a)}{x - a}\right](x - a) = f'(a) \cdot 0 = 0$. So $\lim_{x \to a} f(x) = f(a)$. $\blacksquare$

The converse is false: continuity does not imply differentiability. The classic counter-example is $f(x) = |x|$, continuous everywhere but not differentiable at $0$.

Derivative as a function

If $f$ is differentiable at every $x$ in an interval, the function $x \mapsto f'(x)$ is the derivative function, denoted $f'$ or $\dfrac{df}{dx}$.

Computing from first principles

The derivative of $f(x) = x^2$ at $a$:

$f'(a) = \lim_{h \to 0} \dfrac{(a + h)^2 - a^2}{h} = \lim_{h \to 0}\dfrac{2ah + h^2}{h} = \lim_{h \to 0}(2a + h) = 2a$.

So $f'(x) = 2x$ for every $x$.

When differentiability fails

A continuous function $f$ can fail to be differentiable at $a$ for several reasons:

1. Corner (sharp point): left and right derivatives exist but differ. E.g., $|x|$ at $0$.
2. Vertical tangent: derivative becomes infinite. E.g., $\sqrt[3]{x}$ at $0$.
3. Cusp: vertical tangent with the function turning around. E.g., $x^{2/3}$ at $0$.
4. Oscillation: derivative limit doesn't exist (rare in school problems).

Worked examples

Example 1. Find $f'(x)$ from first principles for $f(x) = x^3$.

$\lim_{h \to 0}\dfrac{(x + h)^3 - x^3}{h} = \lim_{h \to 0}\dfrac{3x^2 h + 3 x h^2 + h^3}{h} = 3x^2$.

Example 2. Show $f(x) = |x|$ is not differentiable at $0$.

Right derivative: $\lim_{h \to 0^+}\dfrac{|h|}{h} = 1$. Left derivative: $\lim_{h \to 0^-}\dfrac{|h|}{h} = -1$. Unequal; not differentiable.

Example 3. Show $f(x) = \sqrt[3]{x}$ is not differentiable at $0$.

$f'(0) = \lim_{h \to 0}\dfrac{h^{1/3}}{h} = \lim_{h \to 0} h^{-2/3} = \infty$. Limit doesn't exist (as a finite number).

Example 4. Find $f'(x)$ from first principles for $f(x) = \sin x$.

$\lim_{h \to 0}\dfrac{\sin(x + h) - \sin x}{h} = \lim_{h \to 0}\dfrac{2 \cos(x + h/2) \sin(h/2)}{h} = \cos x \cdot \lim_{h \to 0}\dfrac{\sin(h/2)}{h/2} \cdot 1 = \cos x \cdot 1 = \cos x$.

Example 5. Where is $f(x) = |x|^3$ differentiable? Compute the derivative.

$f(x) = x^3$ if $x \ge 0$, and $f(x) = -x^3$ if $x < 0$. From either side at $0$: right derivative $\lim_{h \to 0^+} h^2 = 0$. Left derivative $\lim_{h \to 0^-}(-h^2) = 0$. So $f'(0) = 0$ , differentiable. Elsewhere $f'(x) = 3x|x|$.

Example 6. Discuss differentiability of $f(x) = \begin{cases} x^2 \sin(1/x), & x \neq 0 \\ 0, & x = 0 \end{cases}$ at $x = 0$.

$f'(0) = \lim_{h \to 0}\dfrac{h^2 \sin(1/h) - 0}{h} = \lim_{h \to 0} h \sin(1/h) = 0$ (by squeeze). So differentiable at $0$ with $f'(0) = 0$. Note the derivative function is discontinuous at $0$.

Try it yourself

1. From first principles, find $f'(x)$ for $f(x) = 1/x$.
2. From first principles, find $f'(x)$ for $f(x) = \sqrt{x}$.
3. Show $f(x) = |x - 1|$ is not differentiable at $x = 1$.
4. Is $f(x) = x|x|$ differentiable at $0$?
5. Find $f'(x)$ from first principles for $f(x) = e^x$.
6. Is the derivative of $|x|$ even or odd? Sketch its graph.
7. Compute $f'(0)$ if $f(x) = x \sin(1/x)$ for $x \neq 0$ and $f(0) = 0$.
8. Investigate differentiability of $f(x) = \cos|x|$ everywhere.
9. Find conditions on $a, b$ so $f(x) = \begin{cases} ax + b, & x < 1 \\ x^2, & x \ge 1 \end{cases}$ is differentiable at $x = 1$.
10. Show: a function differentiable on $[a, b]$ is also continuous there.
11. Is $f(x) = \tan x$ differentiable on $(-\pi/2, \pi/2)$?
12. Where is $f(x) = \min(x, x^2)$ non-differentiable?
13. Investigate differentiability of $f(x) = |\sin x|$.
14. Show: if $f$ is differentiable at $a$ and $f(a) = 0$, then $\lim_{x \to a}\dfrac{f(x)}{x - a} = f'(a)$.

Pitfalls / Tricks

• Differentiability is stronger than continuity. The implication is one-way.
• Corners ($|x|$ type), cusps, and vertical tangents are the three school-relevant ways to fail differentiability.
• Always check both one-sided derivatives when the function is defined piecewise.
• First-principles computations require careful algebraic manipulation; check limits using trig identities or rationalisation as needed.

Next, the rules of differentiation.