Continuity and Differentiability — The Foundations

Continuity at $c$: $\lim_{x \to c} f(x) = f(c)$ — left hand limit, right hand limit and value all present and equal. Piecewise functions can only misbehave at their joints; check LHL, RHL and value at each joint separately. Sums, products, quotients (denominator nonzero) and composites of continuous functions stay continuous; polynomials, rational functions, $\sin$, $\cos$, $e^x$, $\log x$ and $\vert x \vert$ are continuous throughout their domains, while $[x]$ jumps at every integer.

Continuity-constant questions (find $k$): equate the two sides at the joint. When the piece is a limit form, use the standard limits $\lim_{x\to 0}\frac{\sin kx}{x} = k$, $\lim_{x\to 0}\frac{1 - \cos kx}{x^2} = \frac{k^2}{2}$, $\lim_{x\to 0}\frac{e^{kx} - 1}{x} = k$.

Differentiability: $f'(c) = \lim_{h \to 0}\frac{f(c+h) - f(c)}{h}$, when the limit exists. Every differentiable function is continuous, but not conversely — $\vert x \vert$ is continuous at $0$ yet has LHD $-1$ and RHD $+1$: a corner. Every $\vert x - a \vert$ carries one corner at $a$, and $[x]$ fails at integers by failing continuity first.

The Differentiation Toolkit

The table: $\frac{d}{dx}x^n = nx^{n-1}$; $\sin \to \cos$; $\cos \to -\sin$; $\tan \to \sec^2$; $\sin^{-1}x \to \frac{1}{\sqrt{1-x^2}}$; $\cos^{-1}x \to -\frac{1}{\sqrt{1-x^2}}$; $\tan^{-1}x \to \frac{1}{1+x^2}$; $e^x \to e^x$; $\log x \to \frac{1}{x}$; $a^x \to a^x\log a$.

Chain rule — the engine of everything: $\frac{d}{dx}v(u(x)) = v'(u(x))\cdot u'(x)$, one factor per layer, and forgetting the inner factor is the most common error in the chapter.

Implicit differentiation: differentiate the relation as it stands; every appearance of $y$ triggers a $\frac{dy}{dx}$ by the chain rule; collect and solve — the answer may contain both $x$ and $y$.

Inverse-trig simplification first: arguments like $\frac{2x}{1+x^2}$ collapse under $x = \tan\theta$ to $2\tan^{-1}x$; differentiating the collapsed form saves a page.

Logarithmic differentiation — for variable-to-variable powers $u(x)^{v(x)}$ (both positive where needed): take $\log y = v\log u$, differentiate, multiply back by $y$. Classify before differentiating: $x^a$ power rule, $a^x$ exponential rule, $x^x$ logarithmic differentiation, $a^a$ constant.

Parametric forms: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ (with $\frac{dx}{dt} \neq 0$) — $y$'s rate on top. Standard collapses: $\frac{\sin\theta}{1+\cos\theta} = \tan\frac{\theta}{2}$ and $\frac{\sin\theta}{1-\cos\theta} = \cot\frac{\theta}{2}$.

Second order: $\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right)$. For relation proofs — the flagship 5-marker — use the clear-then-differentiate trick: after the first derivative, clear radicals or denominators, then differentiate the cleaned equation implicitly and substitute back.

Worked Examples — Continuity and Differentiability

Example 1 — A joint that fails

Find all points of discontinuity of $f(x) = \begin{cases} 2x + 3, & x \leq 2 \\ 2x - 3, & x > 2 \end{cases}$.

Step 1 — away from the joint: each piece is a polynomial — continuous.

Step 2 — at $x = 2$: LHL $= 7$, RHL $= 1$ — unequal.

Answer: discontinuous only at $x = 2$ — piecewise functions are innocent except at their joints.

Example 2 — A continuity constant

Find $k$ so that $f(x) = \begin{cases} kx^2, & x \leq 2 \\ 3, & x > 2 \end{cases}$ is continuous at $x = 2$.

Step 1 — equate the sides: LHL $= 4k$ must equal RHL $= 3$ (and $f(2) = 4k$ then agrees automatically).

Answer: $k = \dfrac{3}{4}$.

Example 3 — Continuous but not differentiable

Show that $f(x) = \vert x - 5 \vert$ is continuous but not differentiable at $x = 5$.

Step 1 — continuity: LHL $=$ RHL $= 0 = f(5)$. ✓

Step 2 — one-sided derivatives: LHD $= \lim_{h \to 0^-}\frac{\vert h \vert}{h} = -1$, RHD $= +1$ — unequal.

Answer: continuous with a corner — the standing counterexample to "continuous implies differentiable".

Example 4 — Chain rule, twice over

Differentiate (i) $\sin(x^2 + 5)$, (ii) $(2x + 1)^3$.

Step 1 — (i): outer $\sin$, inner $x^2 + 5$: derivative $2x\cos(x^2 + 5)$.

Step 2 — (ii): power rule outside, inner derivative $2$: $3(2x+1)^2 \cdot 2 = 6(2x+1)^2$.

Answer: as computed — every chain-rule answer carries the inner derivative as a visible factor.

Worked Examples — Implicit, Inverse-Trig, Logarithmic

Example 5 — Implicit with a chain inside

Find $\dfrac{dy}{dx}$ if $y + \sin y = \cos x$.

Step 1 — differentiate through: $\frac{dy}{dx} + \cos y\frac{dy}{dx} = -\sin x$.

Step 2 — factor and solve: $$\frac{dy}{dx} = -\frac{\sin x}{1 + \cos y}, \qquad y \neq (2n+1)\pi$$

Answer: as displayed — the exclusion keeps the denominator alive, and answers in $x$ and $y$ are perfectly acceptable.

Example 6 — Collapse before differentiating

Differentiate $y = \sin^{-1}\left(\dfrac{2x}{1 + x^2}\right)$ for $\vert x \vert < 1$.

Step 1 — substitute $x = \tan\theta$: the argument becomes $\sin 2\theta$, so $y = 2\theta = 2\tan^{-1}x$ on this interval.

Step 2 — differentiate the collapsed form: $\frac{2}{1 + x^2}$.

Answer: $\dfrac{2}{1 + x^2}$ — three lines instead of a quotient-rule page.

Example 7 — The classification table in action

Differentiate $y = x^x$, $x > 0$, and state the derivative of $a^x$ for constant $a > 0$.

Step 1 — log the variable-variable power: $\log y = x\log x$, so $\frac{1}{y}y' = \log x + 1$.

Step 2 — multiply back: $\frac{d}{dx}x^x = x^x(1 + \log x)$.

Step 3 — the constant-base cousin: $\frac{d}{dx}a^x = a^x\log a$.

Answer: as displayed — and $x^a$ takes the plain power rule: three shapes, three different tools.

Example 8 — Full logarithmic differentiation

Differentiate $x^{\sin x}$, $x > 0$, w.r.t. $x$.

Step 1 — log: $\log y = \sin x\log x$.

Step 2 — product rule: $\frac{1}{y}y' = \cos x\log x + \frac{\sin x}{x}$.

Answer: $\dfrac{dy}{dx} = x^{\sin x}\left(\cos x\log x + \dfrac{\sin x}{x}\right)$ — the original function always rides out front.

Example 9 — Exponential and logarithm composites

Differentiate (i) $e^{x^3}$, (ii) $\log(\log x)$ for $x > 1$.

Step 1 — (i): the exponential reproduces itself, times the inner derivative: $3x^2e^{x^3}$.

Step 2 — (ii): two chain links: $\frac{1}{\log x}\cdot\frac{1}{x} = \frac{1}{x\log x}$.

Answer: as computed — the derivative pair $e^x \to e^x$, $\log x \to \frac{1}{x}$ plus the chain rule covers the whole family.

Worked Examples — Parametric and Second Order

Example 10 — The parabola

Find $\dfrac{dy}{dx}$ if $x = at^2$, $y = 2at$.

Step 1 — differentiate each w.r.t. $t$ and divide: $\frac{2a}{2at} = \frac{1}{t}$.

Answer: $\dfrac{1}{t}$ — answers in terms of the parameter are complete answers.

Example 11 — The cycloid with the half-angle finish

Find $\dfrac{dy}{dx}$ if $x = a(\theta + \sin\theta)$, $y = a(1 - \cos\theta)$.

Step 1 — divide the rates: $\dfrac{a\sin\theta}{a(1 + \cos\theta)}$.

Step 2 — half-angle collapse: $\dfrac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\cos^2\frac{\theta}{2}} = \tan\dfrac{\theta}{2}$.

Answer: $\tan\dfrac{\theta}{2}$ — the unsimplified quotient loses the presentation mark.

Example 12 — The sine-cosine relation

If $y = A\sin x + B\cos x$, prove $\dfrac{d^2y}{dx^2} + y = 0$.

Step 1 — differentiate twice: $y'' = -A\sin x - B\cos x = -y$.

Answer: $y'' + y = 0$ for every choice of constants. ∎

Example 13 — The exponential relation

If $y = 3e^{2x} + 2e^{3x}$, prove $y'' - 5y' + 6y = 0$.

Step 1 — the two derivatives: $y' = 6e^{2x} + 6e^{3x}$, $y'' = 12e^{2x} + 18e^{3x}$.

Step 2 — substitute: coefficients of $e^{2x}$: $12 - 30 + 18 = 0$; of $e^{3x}$: $18 - 30 + 12 = 0$.

Answer: proved. ∎ (Pattern: $e^{kx}$ passes through as $k^2 - 5k + 6 = (k-2)(k-3)$, zero at the exponents present.)

Example 14 — Clear-then-differentiate

If $y = \sin^{-1}x$, show that $(1 - x^2)y'' - xy' = 0$.

Step 1 — first derivative, then clear the radical: $y' = \frac{1}{\sqrt{1 - x^2}}$, so $\sqrt{1 - x^2}\,y' = 1$.

Step 2 — differentiate the cleaned equation: $\sqrt{1 - x^2}\,y'' - \frac{x}{\sqrt{1 - x^2}}y' = 0$.

Step 3 — multiply by $\sqrt{1 - x^2}$: $(1 - x^2)y'' - xy' = 0$. ∎

Answer: proved — the clean-up before the second differentiation is the whole trick, and it recurs in every relation proof of this type.