Continuity and Differentiability

Written and reviewed by Devendra Dubey, Mathematics faculty at UniDrill (M.Sc. Mathematics, Osmania University · 10 years teaching Mathematics). Based on NCERT Mathematics Class XII, Chapter 5.

Can a speedometer read a sharp corner?

Draw a graph without lifting your pen and you have drawn a continuous function. Now imagine driving along it: the slope of the graph at each moment is your speed, and the slope is exactly what a derivative measures. A car can speed up, slow down and even change direction smoothly, but it cannot have two different speeds at the same instant. A graph with a sharp corner is a graph with two slopes at one point. That is the whole story of this chapter: continuity asks whether the graph is unbroken, differentiability asks whether it is also smooth, and the second is strictly harder to satisfy than the first.

Continuity: no breaks in the graph

A function $f$ is continuous at a point $c$ if three things hold together: the left hand limit exists, the right hand limit exists, and both equal the value $f(c)$.

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

Break any one of the three and continuity fails. A function is continuous if it is continuous at every point of its domain. On a closed interval $[a, b]$, the end points use one sided limits: $\lim_{x \to a^+} f(x) = f(a)$ and $\lim_{x \to b^-} f(x) = f(b)$.

There are only three ways to fail, and you can name each one by looking at the limits.

The greatest integer function $[x]$ is a staircase. At every integer it jumps by $1$, so it is discontinuous at each integer and continuous everywhere else. At $x = 2$, for instance, the left limit is $1$, the right limit is $2$ and $f(2) = 2$.

Building blocks. Constants, $x$, polynomials, rational functions (where the denominator is not $0$), $\lvert x \rvert$, $\sin x$, $\cos x$, $\tan x$ (where defined) and $e^x$ are all continuous on their domains. Two theorems let you combine them. If $f$ and $g$ are continuous at $c$, then so are $f \pm g$, $f \cdot g$ and $f/g$ (when $g(c) \ne 0$). If $g$ is continuous at $c$ and $f$ is continuous at $g(c)$, then the composite $f \circ g$ is continuous at $c$. Most "is this continuous?" questions are answered by recognising pieces and combining them.

Worked example: finding a constant at a junction. Find $k$ so that $f(x) = kx^2$ for $x \le 2$ and $f(x) = 3$ for $x > 2$ is continuous at $x = 2$.

Common wrong step. Using the wrong piece for the limit, for example substituting $x = 2$ into $3$ for the left limit, or squaring the wrong number to get $2k$ instead of $4k$. Always ask which formula governs each side.

Worked example: removing a hole. Find $k$ so that $f(x) = \dfrac{x^2 - 9}{x - 3}$ for $x \ne 3$ and $f(3) = k$ is continuous at $x = 3$.

Common wrong step. Giving up because "$\dfrac{0}{0}$ is undefined", or putting $k = 0$. The limit only asks what happens near $3$, where the expression is simply $x + 3$.

Pause and predict: is the greatest integer function $[x]$ continuous at $x = 2.5$? And at $x = 2$?

Check your answer: At $2.5$ yes: near it the function is constant at $2$. At $x = 2$ no: the left limit is $1$ and the right limit is $2$, so the limits disagree.

Beyond the textbook. For the exam: $[x]$ is the standard example of a function with a jump at every integer. In reality: step patterns are everywhere. Postal and courier charges that rise in fixed steps with weight, and income tax slabs applied to a band of income, behave like step functions: nothing changes and then, at one point, the value jumps. Why it matters: such real quantities are discontinuous exactly at the thresholds, and the limit test from this section finds them.

Differentiability: a graph with one slope

The derivative of $f$ at $c$ is the slope of the tangent there:

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

provided the limit exists. Approaching from the left gives the left hand derivative and from the right the right hand derivative. The function is differentiable at $c$ exactly when both exist, are finite and are equal.

Differentiable implies continuous. Suppose $f'(c)$ exists. Write $f(c + h) - f(c) = \dfrac{f(c + h) - f(c)}{h} \cdot h$. As $h \to 0$, the first factor tends to $f'(c)$ and the second to $0$, so the product tends to $0$. That says $f(c + h) \to f(c)$, which is continuity.

The converse is false. Take $f(x) = \lvert x \rvert$ at $x = 0$. It is continuous there, but look at the slopes. For $h > 0$, $\dfrac{\lvert h \rvert - 0}{h} = \dfrac{h}{h} = 1$. For $h < 0$, $\dfrac{\lvert h \rvert - 0}{h} = \dfrac{-h}{h} = -1$. The left hand derivative is $-1$ and the right hand derivative is $1$. The graph has a corner at $0$, so there is no single tangent slope.

Pause and predict: a function has a jump at $x = 4$. Can it be differentiable there?

Check your answer: No. Differentiable implies continuous, so a function that is not continuous at $4$ cannot be differentiable at $4$. This is the same theorem read backwards.

Beyond the textbook. For the exam: continuity does not imply differentiability, and $\lvert x \rvert$ at $0$ is the example to quote. In reality: the gap is far wider than a few corners. In 1872 Weierstrass gave a function that is continuous at every point of the real line and differentiable at none: a graph with no break and no smooth stretch anywhere. Why it matters: it shows that "unbroken" and "smooth" are genuinely different ideas, not one idea in two vocabularies.

Rules of differentiation

If $u$ and $v$ are differentiable, so are their combinations:

$$(u \pm v)' = u' \pm v', \qquad (uv)' = u'v + uv', \qquad \left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2} \ \ (v \ne 0)$$

For the quotient rule the numerator order matters: bottom times derivative of top, minus top times derivative of bottom. The denominator is squared.

The chain rule handles composites. If $y = v(t)$ and $t = u(x)$, then

$$\dfrac{dy}{dx} = \dfrac{dy}{dt} \cdot \dfrac{dt}{dx}$$

Differentiate the outer function, keeping the inner one intact, then multiply by the derivative of the inner function. For three layers there are three factors. So $\dfrac{d}{dx}\sin(x^2) = \cos(x^2) \cdot 2x$.

Where the inverse trigonometric derivatives come from. Let $y = \sin^{-1}x$, so $x = \sin y$ with $y$ in $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$. Differentiating both sides with respect to $x$ gives $1 = \cos y \cdot \dfrac{dy}{dx}$, so $\dfrac{dy}{dx} = \dfrac{1}{\cos y}$. On this branch $\cos y \ge 0$, so $\cos y = +\sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}$. That is why the answer is positive, and why it is valid only on the open interval $(-1, 1)$: at $x = \pm 1$ we have $\cos y = 0$. The same method with $x = \cos y$ on $[0, \pi]$, where $\sin y \ge 0$, gives the negative sign for $\cos^{-1}$. For $x = \tan y$ we get $1 = \sec^2 y \cdot \dfrac{dy}{dx}$ and $\sec^2 y = 1 + x^2$.

Here $\log x$ means the natural logarithm, with base $e$, throughout this chapter. The derivative of $a^x$ comes from taking logarithms: if $y = a^x$ then $\log y = x\log a$, so $\dfrac{1}{y}\dfrac{dy}{dx} = \log a$ and $\dfrac{dy}{dx} = a^x\log a$. And $\log_a x = \dfrac{\log x}{\log a}$, which gives $\dfrac{d}{dx}\log_a x = \dfrac{1}{x\log a}$.

Worked example: chain rule with an inverse function. Differentiate $y = \sin^{-1}(2x - 1)$.

Common wrong step. Writing $\dfrac{1}{\sqrt{1 - (2x - 1)^2}}$ and forgetting the factor $2$ from the inner function. Whenever the input is not plain $x$, the chain rule adds a factor.

Implicit differentiation

When $x$ and $y$ are tangled in one equation and you cannot easily solve for $y$, differentiate both sides with respect to $x$, treating $y$ as a function of $x$. Every time you differentiate a term in $y$, the chain rule leaves a $\dfrac{dy}{dx}$ behind. Then collect the $\dfrac{dy}{dx}$ terms and solve.

Worked example: an implicit curve. Find $\dfrac{dy}{dx}$ if $x^2 + xy + y^2 = 100$.

Common wrong step. Treating $y$ as a constant, so that $xy$ becomes just $y$ and $y^2$ becomes $2y$ with no $\dfrac{dy}{dx}$. Remember that $y$ is a function of $x$.

Logarithmic differentiation

The power rule fails for $x^x$ because the exponent moves, and the exponential rule fails because the base moves. The way out is to take logarithms first. For $y = [u(x)]^{v(x)}$ with $u > 0$, write $\log y = v\log u$ and differentiate both sides:

$$\dfrac{dy}{dx} = u^v\left(v'\log u + \dfrac{v\,u'}{u}\right)$$

Worked example: the variable power. Find $\dfrac{dy}{dx}$ for $y = x^x$, $x > 0$.

Common wrong step. Using the power rule alone to get $x \cdot x^{x-1} = x^x$, or the exponential rule alone to get $x^x\log x$. Look closely: these two wrong answers are exactly the two halves of the right one. The true derivative is their sum, because the exponent and the base are both changing.

Parametric and second order derivatives

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

$$\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{g'(t)}{f'(t)}, \quad f'(t) \ne 0$$

For example, with $x = at^2$ and $y = 2at$ we have $\dfrac{dx}{dt} = 2at$ and $\dfrac{dy}{dt} = 2a$, so $\dfrac{dy}{dx} = \dfrac{2a}{2at} = \dfrac{1}{t}$. The answer is a ratio of two derivatives with respect to $t$, never the derivative of $y$ with respect to $t$ alone.

The second derivative is the derivative of the derivative: $\dfrac{d^2y}{dx^2} = \dfrac{d}{dx}\left(\dfrac{dy}{dx}\right)$, also written $f''(x)$ or $y''$. It measures how fast the slope itself is changing, just as acceleration measures how fast speed changes.

Worked example: a second derivative. Find $\dfrac{d^2y}{dx^2}$ for $y = x^3 + \tan x$.

Common wrong step. Differentiating $\sec^2 x$ as $2\sec x$ and stopping, which skips the inner derivative $\sec x\tan x$.

Beyond the textbook. For the exam: $\log x$ in this chapter is the natural logarithm, so $\dfrac{d}{dx}\log x = \dfrac{1}{x}$. In reality: calculators have two keys, log for base $10$ and ln for base $e$, and that causes confusion. Most programming languages (C, Python, JavaScript) use "log" for the natural logarithm, just as this chapter does. Why it matters: the number $e$ is chosen as the base because it is the one for which $e^x$ is its own derivative, with no stray constant. For any other base $a$ the extra factor $\log a$ appears.

Exam traps

Formula sheet

60-second revision

Where this chapter connects

• Previous: Determinants closes the algebra half of Class XII.
• Class XI link: Limits and Derivatives introduces limits and the first principles definition that this chapter builds on.
• Same course: Inverse Trigonometric Functions supplies the branches that decide the sign and domain of the inverse trigonometric derivatives.
• Next: Application of Derivatives uses first and second derivatives for rates, tangents and maxima, and Integrals reverses differentiation.