Differentiability

The derivative of $f$ at a point $c$ of its domain is

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

provided this limit exists — and that proviso is the whole subject of this section. When the limit exists, $f$ is differentiable at $c$; when it exists at every point, the function $f'(x)$ (also written $\frac{dy}{dx}$ or $y'$) is the derivative of $f$, and finding it is differentiation.

The algebra of derivatives from Class 11 carries over: $(u \pm v)' = u' \pm v'$, the product rule $(uv)' = u'v + uv'$, and the quotient rule $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$ where $v \neq 0$; with the standard table $\frac{d}{dx}x^n = nx^{n-1}$, $\frac{d}{dx}\sin x = \cos x$, $\frac{d}{dx}\cos x = -\sin x$, $\frac{d}{dx}\tan x = \sec^2 x$.

The one-way theorem

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

Proof idea: write $f(x) - f(c) = \frac{f(x) - f(c)}{x - c} \cdot (x - c)$; as $x \to c$ the first factor tends to $f'(c)$ and the second to $0$, so $f(x) \to f(c)$. ∎

The converse fails. $f(x) = \vert x \vert$ is continuous everywhere, but at $0$: $$\text{LHD} = \lim_{h \to 0^-}\frac{\vert h \vert - 0}{h} = \frac{-h}{h} = -1, \qquad \text{RHD} = \lim_{h \to 0^+}\frac{h}{h} = 1$$ The one-sided derivatives disagree, so no derivative exists at $0$ — the graph has a corner.

So the hierarchy runs one way only: differentiable $\Rightarrow$ continuous, never backwards. Two standing counterexamples for exams: $\vert x - a \vert$ (corner at $a$) and $[x]$ (not even continuous at integers, hence certainly not differentiable there).

The chain rule

For a composite $f = v \circ u$ — that is, $f(x) = v(u(x))$ — set $t = u(x)$. If $\frac{dt}{dx}$ and $\frac{dv}{dt}$ both exist, then

$$\boxed{\;\frac{df}{dx} = \frac{dv}{dt} \cdot \frac{dt}{dx}\;}$$

— differentiate the outer function at the inner value, then multiply by the derivative of the inner function. It extends to longer chains: for $w \circ v \circ u$, multiply three links.

Worked pattern: $f(x) = (2x + 1)^3$. Outer: cube; inner: $t = 2x + 1$. Then $\frac{df}{dx} = 3t^2 \cdot 2 = 6(2x + 1)^2$ — no expansion of the cube needed. Similarly $\frac{d}{dx}\sin(x^2) = \cos(x^2) \cdot 2x$.

The chain rule is the workhorse of this entire chapter — implicit, logarithmic and parametric differentiation are all the chain rule wearing different costumes.

Solved Examples

Example 1 — Modulus is not differentiable at its corner

Prove that $f(x) = \vert x - 1 \vert$ is not differentiable at $x = 1$.

Step 1 — left hand derivative: for $h < 0$: $\frac{f(1 + h) - f(1)}{h} = \frac{\vert h \vert}{h} = \frac{-h}{h} = -1$.

Step 2 — right hand derivative: for $h > 0$: $\frac{\vert h \vert}{h} = 1$.

Answer: LHD $= -1 \neq 1 =$ RHD, so the defining limit does not exist — not differentiable at $x = 1$ (though perfectly continuous there). Every $\vert x - a \vert$ has exactly this corner at $x = a$.

Example 2 — Chain rule, single link

Differentiate $\sin(x^2 + 5)$ with respect to $x$.

Step 1 — name the layers: outer $\sin$, inner $t = x^2 + 5$.

Step 2 — multiply the links: $\frac{dy}{dx} = \cos(x^2 + 5) \cdot (2x)$.

Answer: $2x\cos(x^2 + 5)$.

Example 3 — Chain rule with a trig inner function

Differentiate $\cos(\sin x)$ with respect to $x$.

Step 1 — layers: outer $\cos$, inner $t = \sin x$.

Step 2 — multiply: $\frac{dy}{dx} = -\sin(\sin x) \cdot \cos x$.

Answer: $-\cos x \sin(\sin x)$ — the inner derivative $\cos x$ must not be forgotten; leaving it off is the classic chain-rule slip.

Example 4 — A three-link chain

Differentiate $\sec\left(\tan\sqrt{x}\right)$ with respect to $x$.

Step 1 — three layers: outermost $\sec$, middle $\tan$, innermost $\sqrt{x}$.

Step 2 — multiply all three links: $$\frac{dy}{dx} = \sec(\tan\sqrt{x})\tan(\tan\sqrt{x}) \cdot \sec^2\sqrt{x} \cdot \frac{1}{2\sqrt{x}}$$

Answer: as displayed — work outside-in, one derivative per layer, and multiply everything.

Example 5 — Square root outer layer

Differentiate $\cos(\sqrt{x})$ with respect to $x$.

Step 1 — layers: outer $\cos$, inner $\sqrt{x}$ with derivative $\frac{1}{2\sqrt{x}}$.

Step 2 — multiply: $\frac{dy}{dx} = -\sin(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}}$.

Answer: $-\dfrac{\sin\sqrt{x}}{2\sqrt{x}}$, valid for $x > 0$.

Example 6 — The greatest integer function fails twice over

Prove that $f(x) = [x]$, $0 < x < 3$, is not differentiable at $x = 1$ and $x = 2$.

Step 1 — the fast route: at $x = 1$ and $x = 2$ the function is not even continuous (unit jumps), and differentiability implies continuity.

Step 2 — conclude by contraposition: not continuous at a point $\Rightarrow$ not differentiable there.

Answer: $f$ is not differentiable at $x = 1$ or $x = 2$. ∎ The one-way theorem used backwards — discontinuity is an instant differentiability disqualifier — saves computing any difference quotients.