Continuity and Differentiability — Class 12 IMO

Continuity

Informally, a function is continuous if you can draw its graph without lifting your pen. The precise definition pins this idea to limits.

Continuity at a point

A function $f$ is continuous at $x=c$ if all three conditions hold:

$f(c)$ is defined,
$\displaystyle\lim_{x\to c} f(x)$ exists (left-hand limit $=$ right-hand limit), and
$\displaystyle\lim_{x\to c} f(x)=f(c)$.

If any one fails, $f$ is discontinuous at $c$. For piecewise functions, the usual exam task is to match the left-hand limit, right-hand limit and function value at the join.

Continuity on an interval

$f$ is continuous on an interval if it is continuous at every point of it. Polynomials, $\sin x$, $\cos x$, $e^{x}$ are continuous everywhere; $\tan x$ is continuous except where $\cos x=0$.

Algebra of continuous functions

If $f$ and $g$ are continuous at $c$, so are $f\pm g$, $fg$, and $\dfrac{f}{g}$ (provided $g(c)\ne0$). Compositions of continuous functions are continuous. This lets you certify most expressions as continuous without epsilon-delta work.

Example 1: Is $f(x)=\begin{cases} x+1, & x\le 2\\ 2x-1, & x>2\end{cases}$ continuous at $x=2$?

$f(2)=2+1=3$. LHL $=\lim_{x\to2^-}(x+1)=3$. RHL $=\lim_{x\to2^+}(2x-1)=3$. All three agree, so $f$ is continuous at $x=2$.

Example 2: Find $k$ so that $f(x)=\begin{cases} kx^2, & x\le 1\\ 4, & x>1\end{cases}$ is continuous at $x=1$.

RHL $=4$ and $f(1)=k$. LHL $=\lim_{x\to1^-}kx^2=k$. For continuity $k=4$.

Example 3: Discuss the continuity of $f(x)=\dfrac{1}{x-3}$.

It is a quotient of continuous functions, continuous wherever the denominator is non-zero, i.e. for all $x\ne3$. At $x=3$ the function is undefined, so it is discontinuous only there.

Example 4: Show $f(x)=|x|$ is continuous at $x=0$.

$f(0)=0$. LHL $=\lim_{x\to0^-}(-x)=0$, RHL $=\lim_{x\to0^+}(x)=0$. All equal $0$, so $|x|$ is continuous at $0$ (though, as the next page shows, not differentiable there).

Quick recap

Continuous at $c$ $\iff$ $f(c)$ defined, $\lim_{x\to c}f(x)$ exists, and the two are equal.
For piecewise functions, equate LHL, RHL and $f(c)$ at each join.
Polynomials, $\sin,\cos,e^{x}$ are continuous everywhere; rational functions wherever the denominator is non-zero.
Sums, products, quotients (non-zero denom) and compositions of continuous functions are continuous.

✓ Quick check

If f(x) = x³ − 3x, then the value of c in (0, 2) where tangent is parallel to chord joining (0, f(0)) and (2, f(2)) is:

f(0)=0, f(2)=8−6=2. Slope = (2−0)/(2−0)=1. f′(x) = 3x² − 3 = 1 → 3x² = 4 → x² = 4/3 → x = 2/√3 (in (0,2)).

If x² + y² = 25, then y″ at (3, 4) is:

2x + 2y y′ = 0 → y′ = −x/y. At (3,4): y′ = −3/4. y″ = −(y − x y′)/y² = −(4 − 3(−3/4))/16 = −(4 + 9/4)/16 = −(25/4)/16 = −25/64.

Differentiability and the Chain Rule

The derivative of $f$ at $x=c$ is the limit of the average rate of change:

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

provided this limit exists. Geometrically it is the slope of the tangent at $\big(c,f(c)\big)$.

Differentiability implies continuity

If $f$ is differentiable at $c$, then it is continuous at $c$. The converse is false: $f(x)=|x|$ is continuous at $0$ but has a corner there, so $f'(0)$ does not exist (left slope $-1$, right slope $+1$). Corners and sharp points break differentiability.

Standard derivatives

$f(x)$	$f'(x)$	$f(x)$	$f'(x)$
$x^{n}$	$nx^{n-1}$	$\sin x$	$\cos x$
$\cos x$	$-\sin x$	$\tan x$	$\sec^2 x$
$e^{x}$	$e^{x}$	$\ln x$	$\tfrac1x$

The chain rule

For a composite $y=f(g(x))$,

$$\frac{dy}{dx}=f'\big(g(x)\big)\cdot g'(x).$$

Differentiate the outer function, keep the inside, then multiply by the derivative of the inside. Combined with the product rule $(uv)'=u'v+uv'$ and quotient rule, this handles almost every function in the chapter.

Example 1: Differentiate $y=(3x^2+1)^5$.

Chain rule with outer $u^5$, inner $u=3x^2+1$: $\dfrac{dy}{dx}=5(3x^2+1)^4\cdot 6x = 30x(3x^2+1)^4.$

Example 2: Differentiate $y=\sin(x^2)$.

$\dfrac{dy}{dx}=\cos(x^2)\cdot 2x = 2x\cos(x^2).$

Example 3: Differentiate $y=x^2 e^{x}$.

Product rule: $\dfrac{dy}{dx}=2x\,e^{x}+x^2 e^{x}=e^{x}(x^2+2x).$

Example 4: Show $f(x)=|x|$ is not differentiable at $x=0$.

The left-hand derivative is $\lim_{h\to0^-}\dfrac{|h|}{h}=-1$ and the right-hand derivative is $\lim_{h\to0^+}\dfrac{|h|}{h}=+1$. Since $-1\ne+1$, $f'(0)$ does not exist.

Quick recap

$f'(c)=\lim_{h\to0}\dfrac{f(c+h)-f(c)}{h}$ = slope of the tangent.
Differentiable $\Rightarrow$ continuous; continuous does not imply differentiable (e.g. $|x|$ at $0$).
Chain rule: $\dfrac{d}{dx}f(g(x))=f'(g(x))\,g'(x)$.
Product rule $(uv)'=u'v+uv'$; quotient rule $\left(\tfrac uv\right)'=\dfrac{u'v-uv'}{v^2}$.

✓ Quick check

The derivative of sin⁻¹x with respect to x is:

This is a standard formula: d/dx (sin⁻¹x) = 1 / √(1 − x²).

Evaluate the derivative of (tan⁻¹x + cot⁻¹x) with respect to x.

tan⁻¹x + cot⁻¹x = π/2 (a constant) for all real x. The derivative of a constant is 0.

Logarithmic, Implicit and Parametric

Some relations are not given as $y=f(x)$. Three techniques extend differentiation to these cases.

Implicit differentiation

When $x$ and $y$ are tangled in one equation, differentiate both sides with respect to $x$, treating $y$ as a function of $x$ (so $\tfrac{d}{dx}y^n = ny^{n-1}\tfrac{dy}{dx}$), then solve for $\tfrac{dy}{dx}$.

Logarithmic differentiation

For products/quotients of many factors, or for a variable raised to a variable power such as $y=x^{x}$, take $\ln$ of both sides first. This converts powers into products and products into sums, which are far easier to differentiate. For $y=u(x)^{v(x)}$ this is the only elementary route.

Parametric differentiation

If $x=f(t)$ and $y=g(t)$, then

$$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{g'(t)}{f'(t)},\qquad f'(t)\ne0.$$

Second-order derivatives

Differentiating $\tfrac{dy}{dx}$ again gives $\tfrac{d^2y}{dx^2}$, the rate of change of the slope — needed for concavity and the second-derivative test in the next chapter.

Example 1: Find $\dfrac{dy}{dx}$ if $x^2+y^2=25$.

Differentiate implicitly: $2x+2y\dfrac{dy}{dx}=0 \Rightarrow \dfrac{dy}{dx}=-\dfrac{x}{y}.$

Example 2: Differentiate $y=x^{x}$ ($x>0$).

Take logs: $\ln y=x\ln x$. Differentiate: $\dfrac{1}{y}\dfrac{dy}{dx}=\ln x+1$. So $\dfrac{dy}{dx}=x^{x}(1+\ln x).$

Example 3: If $x=at^2,\ y=2at$, find $\dfrac{dy}{dx}$.

$\dfrac{dx}{dt}=2at,\ \dfrac{dy}{dt}=2a$. Hence $\dfrac{dy}{dx}=\dfrac{2a}{2at}=\dfrac{1}{t}.$

Example 4: Find $\dfrac{d^2y}{dx^2}$ for $y=x^3$.

$\dfrac{dy}{dx}=3x^2$, then $\dfrac{d^2y}{dx^2}=6x.$

Quick recap

Implicit: differentiate both sides w.r.t. $x$ (use chain rule on $y$), then solve for $\tfrac{dy}{dx}$.
Logarithmic differentiation is essential for $y=u(x)^{v(x)}$; take $\ln$ first.
Parametric: $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$.
Second derivative $\tfrac{d^2y}{dx^2}$ is the derivative of the first derivative.

✓ Quick check

The area of a circular oil spill in Mumbai harbour is expanding. If A = πr², the rate of change of area with respect to its radius r when r = 10 m is:

dA/dr = 2πr. When r = 10 m, the rate is 2π(10) = 20π square meters per meter.

A train from New Delhi to Agra follows a distance-time curve s(t) = t³ − 6t² + 15t + 10 (s in km, t in hours). What is its acceleration at t = 2 hours?

Velocity v(t) = s'(t) = 3t² − 12t + 15. Acceleration a(t) = v'(t) = 6t − 12. At t = 2, a(2) = 12 − 12 = 0.