Continuity and Differentiability

1. Check this before you revise anything

Rolle's Theorem and the Mean Value Theorem are not in this chapter — a full-text search turns up zero mentions of either. Older editions had a dedicated final section proving both, with their own exercise (5.8) of classic geometric-interpretation questions. The current book stops at second-order derivatives (Section 5.7) and moves straight to the Miscellaneous Exercise. This matches the syllabus line itself, which lists "continuity and differentiability, chain rule... second order derivatives" and nothing about MVT.

The old stub taught Rolle's/MVT as a full section and collapsed the chapter's seven real exercises into one invented 45-question group with no solutions file behind it. This rebuild removes MVT entirely and replaces the fake exercise count with what's actually here.

The book has Exercise 5.1 (34 questions, continuity), Exercise 5.2 (10, chain rule and non-differentiability), Exercise 5.3 (15, implicit and inverse-trig derivatives), Exercise 5.4 (10, exponential/logarithmic derivatives), Exercise 5.5 (18, logarithmic differentiation), Exercise 5.6 (11, parametric forms), Exercise 5.7 (11, second-order derivatives), and a Miscellaneous Exercise (22).

131 questions in total — the largest chapter in this project so far.

2. What this chapter covers

3. Continuity and differentiability

$f$ is continuous at $x=a$ if ${\lim }_{x\to a}f(x)=f(a)$ — which unpacks to three separate things all being true and equal: the left-hand limit, the right-hand limit, and the function's actual value there. For a piecewise function, this is where almost every exercise question in this chapter lives: compute LHL, RHL, and $f(a)$ at the join point, and check all three match.

$f$ is differentiable at $x=a$ if ${\lim }_{h\to 0}\frac{f(a+h)-f(a)}{h}$ exists — meaning the left-hand and right-hand versions of this limit must agree. Differentiability is the stronger condition: differentiable always implies continuous, but not the reverse. $f(x)=|x|$ is the standard counterexample — continuous everywhere, but not differentiable at $x=0$, where the left-hand derivative is $-1$ and the right-hand derivative is $1$.

The three conditions, written out. Continuity at $x=a$ fails if any one of these fails:

1. $f(a)$ is defined (the point is in the domain).
2. ${\lim }_{x\to a}f(x)$ exists, which requires LHL $=$ RHL.
3. The limit equals the value: ${\lim }_{x\to a}f(x)=f(a)$.

A function can clear the first two and still fail the third — a removable discontinuity, where the graph has a hole that the defined value sits away from.

Continuity on an interval. $f$ is continuous on $[a,b]$ if it is continuous at every interior point, right-continuous at $a$, and left-continuous at $b$. The endpoints only need the one-sided condition, since the function is not defined beyond them.

Algebra of continuous functions (5.2.1). If $f$ and $g$ are continuous at $a$, then so are $f+g$, $f-g$, and $f\cdot g$; and $f/g$ is continuous at $a$ provided $g(a)\ne 0$. The composition $g\circ f$ is continuous wherever the pieces are.

This is what licenses the standard shortcut: every polynomial is continuous everywhere, and every rational function is continuous except where its denominator vanishes. Sine, cosine and the exponential are continuous on all of $\mathbb{R}$.

Where the exercise questions actually live. Almost all of Exercise 5.1 is a piecewise function with an unknown constant $k$, asking for the value that makes it continuous. The method is fixed: compute LHL and RHL at the join, set them equal to $f(a)$, and solve for $k$.

Why differentiability implies continuity. If $f^{\prime }(a)$ exists then

${\lim }_{x\to a}\left[f(x)-f(a)\right]={\lim }_{x\to a}\frac{f(x)-f(a)}{x-a}\cdot (x-a)=f^{\prime }(a)\cdot 0=0$

so ${\lim }_{x\to a}f(x)=f(a)$. The converse fails at any corner, where the two one-sided derivatives disagree.

4. The differentiation toolkit

Beyond the ordinary chain rule, this chapter builds three specific techniques:

Implicit differentiation. When $y$ isn't isolated (like $x^2+xy+y^2=100$), differentiate both sides with respect to $x$, applying the chain rule to every $y$-term (so $\frac{d}{dx}(y^2)=2y\frac{dy}{dx}$), then solve algebraically for $\frac{dy}{dx}$.

Logarithmic differentiation. For $y=[u(x){]}^{v(x)}$ — a variable base and a variable exponent — ordinary power and exponential rules don't apply directly. Take $\log$ of both sides first: $\log y=v(x)\log [u(x)]$, differentiate implicitly, then multiply back through by $y$:

$\frac{dy}{dx}=y\left[\frac{v(x)}{u(x)}\cdot u^{\prime }(x)+v^{\prime }(x)\cdot \log [u(x)]\right]$

Derivatives of composite functions (5.3.1) — the chain rule. If $y=f(u)$ and $u=g(x)$ then:

$\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}$

Every other technique in this section is an application or extension of it, and for nested compositions it simply repeats one layer at a time.

Derivatives of inverse trigonometric functions (5.3.3). These are quoted constantly and worth memorising outright:

$\frac{d}{dx}{\sin }^{-1}x=\frac{1}{\sqrt{1-x^2}},\frac{d}{dx}{\cos }^{-1}x=\frac{-1}{\sqrt{1-x^2}},\frac{d}{dx}{\tan }^{-1}x=\frac{1}{1+x^2}$

$\frac{d}{dx}{\cot }^{-1}x=\frac{-1}{1+x^2},\frac{d}{dx}{\sec }^{-1}x=\frac{1}{|x|\sqrt{x^2-1}},\frac{d}{dx}{\text{cosec}}^{-1}x=\frac{-1}{|x|\sqrt{x^2-1}}$

Note the pattern: each "co-" function is the negative of its partner. Only the signs differ, which halves what must actually be remembered.

Exponential and logarithmic functions (5.4). The two base results are:

$\frac{d}{dx}{e}^x=e^x,\frac{d}{dx}\log x=\frac{1}{x}$

and with the chain rule, $\frac{d}{dx}{e}^{f(x)}=e^{f(x)}{f}^{\prime }(x)$ and $\frac{d}{dx}\log f(x)=\frac{f^{\prime }(x)}{f(x)}$. For a general base, $\frac{d}{dx}{a}^x=a^x\log a$.

Parametric forms. When $x=f(t)$ and $y=g(t)$ share a third variable rather than one being a direct function of the other:

$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$

Worked, mirroring the textbook's own technique for inverse-trig composites. Find $\frac{dy}{dx}$ for $y={\tan }^{-1}\left(\frac{3x-x^3}{1-3x^2}\right)$. Substitute $x=\tan \theta$: the fraction becomes $\tan 3\theta$ (the triple-angle tangent formula), so $y={\tan }^{-1}(\tan 3\theta )=3\theta =3{\tan }^{-1}x$, giving $\frac{dy}{dx}=\frac{3}{1+x^2}$ — no direct differentiation of the original messy fraction required.

5. Second-order derivatives

$\frac{d^{2}y}{d{x}^2}$ is just $\frac{dy}{dx}$ differentiated again. For an implicit or parametric relationship, this usually means differentiating $\frac{dy}{dx}$ a second time using the same technique (implicit differentiation again, or dividing by $\frac{dx}{dt}$ again for parametric forms) rather than trying to find a closed form for $y$ first.

The parametric trap. For $x=f(t)$, $y=g(t)$, the second derivative is not $\frac{d^{2}y/d{t}^2}{d^{2}x/d{t}^2}$. You must differentiate $\frac{dy}{dx}$ with respect to $t$ and then divide by $\frac{dx}{dt}$ again:

$\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}$

Taking the ratio of second derivatives directly is one of the most reliably wrong answers in the chapter.

Proving a differential relation. Several questions ask you to show something like $(1-x^2)y_2-x{y}_1-a^{2}y=0$, where $y_1$ and $y_2$ denote the first and second derivatives. The route is always the same: compute $y_1$, compute $y_2$, then substitute both into the left-hand side and simplify to zero. There is no need to solve anything.

Summary

• Continuity at $x=a$ requires LHL $=$ RHL $=f(a)$ — check all three explicitly for piecewise functions.
• Differentiable implies continuous, but continuous does not imply differentiable ($|x|$ at $x=0$ is the standard counterexample).
• Implicit differentiation: differentiate both sides, apply the chain rule to every $y$-term, solve for $\frac{dy}{dx}$.
• Logarithmic differentiation is required whenever both the base and the exponent are functions of $x$.
• Parametric forms: $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$.
• Second-order derivatives are found by differentiating $\frac{dy}{dx}$ again, using whichever technique (implicit, parametric) produced it the first time.
• Continuity needs all three: $f(a)$ defined, the limit existing (LHL $=$ RHL), and the two being equal.
• Sums, differences and products of continuous functions are continuous; quotients are, wherever the denominator is non-zero.
• Chain rule: $\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}$ — every other technique here extends it.
• Inverse-trig derivatives come in signed pairs: each "co-" function is the negative of its partner.
• $\frac{d}{dx}{e}^{f(x)}=e^{f(x)}{f}^{\prime }(x)$, $\frac{d}{dx}\log f(x)=\frac{f^{\prime }(x)}{f(x)}$, and $\frac{d}{dx}{a}^x=a^x\log a$.
• For parametric second derivatives, differentiate $\frac{dy}{dx}$ with respect to $t$ and divide by $\frac{dx}{dt}$ again — never take the ratio of the two second derivatives.
• Rolle's Theorem and the Mean Value Theorem are not part of the current edition of this chapter — they don't appear in the text, in any example, or in any exercise.