Why Some Graphs Can Be Drawn Without Lifting Your Pencil

A graph you can draw without lifting your pencil is continuous; one that also has no sharp corners is differentiable. These two ideas decide where calculus can safely be used.

This part covers continuity at a point, the algebra of continuous functions, the chain rule and its link with continuity, and derivatives of inverse trigonometric and implicit functions.

**A function $f$ is continuous at $x=c$ if its left-hand limit, right-hand limit and value $f(c)$ all exist and are equal, and it is continuous on an interval if it is continuous at every point of that interval.

The condition:**

$$\underset{x\to c^{-}}{\lim }f(x)=\underset{x\to c^{+}}{\lim }f(x)=f(c)$$

Worked example — continuous at the join.

$$f(x)=\{\begin{array}{ll}2x+1,& x\le 1\\ 3x,& x>1\end{array}\text{LHL}=3,\text{RHL}=3,f(1)=3$$

All three agree, so $f$ is **continuous at $x=1$.

Worked example — a jump.** For $g(x)=x+2$ when $x\le 2$ and $g(x)=x-2$ when $x>2$, the left-hand limit at $2$ is $4$ but the right-hand limit is $0$, so $g$ is **discontinuous at $x=2$.

Continuous everywhere.** Polynomials, $\sin x$, $\cos x$, $e^x$ and $|x|$ are continuous at every real number.

An everyday example. A water tank's level rising as a tap runs changes continuously, while a parking fee that jumps from ₹20 to ₹40 the moment two hours pass is discontinuous at that instant.

The substance. Continuity is judged only at points in the domain — $\frac{1}{x}$ is continuous, because $x=0$ is not in its domain.

**If $f$ and $g$ are continuous at a point, then so are $f+g$, $f-g$, $fg$ and, where $g\ne 0$, $\frac{f}{g}$; and if $g$ is continuous at $c$ and $f$ is continuous at $g(c)$, the composite $f\circ g$ is continuous at $c$.

The rules** for $f$ and $g$ continuous at $c$:

- $f+g$, $f-g$ and $fg$ are continuous at $c$
- $\frac{f}{g}$ is continuous at $c$ provided $g(c)\ne 0$
- $kf$ is continuous for any constant $k$

Worked example — a quotient. $h(x)=\frac{x^2+1}{x-3}$ is a quotient of polynomials, so it is continuous at every $x\ne 3$.

Worked example — a composite. $\sin (x^2)$ is $f(g(x))$ with $g(x)=x^2$ and $f(u)=\sin u$, both continuous everywhere, so $\sin (x^2)$ is continuous everywhere. Likewise, $|\cos x|$ is continuous for all real $x$.

Worked example — where it fails. $\tan x=\frac{\sin x}{\cos x}$ is continuous except where $\cos x=0$, that is, at $x=\frac{(2n+1)\pi }{2}$ for integers $n$.

An everyday example. A fuel bill is price per litre times litres used; if both change smoothly over a month, the bill changes smoothly too, just as products of continuous functions stay continuous.

The substance. The rules work in one direction only — two discontinuous functions can add up to a continuous one.

The chain rule differentiates a composite function by multiplying the derivative of the outer function, taken at the inner function, by the derivative of the inner function; and a function differentiable at a point must be continuous there, though the reverse need not hold.

The chain rule. If $y=f(u)$ and $u=g(x)$, then

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

Worked example 1. $\frac{d}{dx}\sin (x^2+1)=\cos (x^2+1)\cdot 2x$.

Worked example 2. For $y=(3x+2{)}^5$ at $x=0$:

$$\frac{dy}{dx}=5(3x+2{)}^4\cdot 3=15\cdot 2^4=240$$

Differentiability implies continuity. If $f^{\prime }(c)$ exists, then

$$\underset{x\to c}{\lim }[f(x)-f(c)]=\underset{x\to c}{\lim }\frac{f(x)-f(c)}{x-c}\cdot (x-c)=f^{\prime }(c)\cdot 0=0$$

so $f$ is continuous at $c$.

The converse fails. $f(x)=|x|$ is continuous at $0$, but its left-hand derivative is $-1$ and right-hand derivative is $1$, so it is **not differentiable at $0$ — the graph has a corner.

An everyday example. Fuel used depends on distance, and distance depends on time, so fuel used per hour is fuel per kilometre times kilometres per hour — the chain rule in daily travel.

The substance. A corner or a vertical tangent breaks differentiability** without breaking continuity.

**The derivatives are $\frac{1}{\sqrt{1-x^2}}$ for ${\sin }^{-1}x$, $-\frac{1}{\sqrt{1-x^2}}$ for ${\cos }^{-1}x$ and $\frac{1}{1+x^2}$ for ${\tan }^{-1}x$, and implicit functions are differentiated term by term, treating $y$ as a function of $x$.

Deriving ${\sin }^{-1}x$.** Let $y={\sin }^{-1}x$, so $\sin y=x$. Differentiating both sides,

$$\cos y\frac{dy}{dx}=1\Rightarrow \frac{dy}{dx}=\frac{1}{\cos y}=\frac{1}{\sqrt{1-x^2}}$$

using $\cos y>0$ for $y$ in $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$.

Worked example — with the chain rule.

$$\frac{d}{dx}{\tan }^{-1}(2x)=\frac{1}{1+4x^2}\cdot 2=\frac{2}{1+4x^2}$$

Implicit differentiation. For $x^2+y^2=25$, differentiate both sides with respect to $x$:

$$2x+2y\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=-\frac{x}{y}$$

At the point $(3,4)$ on this circle, the slope is $-\frac{3}{4}$.

An everyday example. A ladder sliding down a wall keeps its length fixed, so the foot's distance and the top's height are linked by an implicit equation like $x^2+y^2=25$.

The substance. **Every $y$ term picks up a factor of $\frac{dy}{dx}$** when differentiated implicitly — forgetting it is the most common slip.