Continuity and Differentiability

The chapter has a clear spine: definitions first, then bigger and bigger classes of functions you can differentiate. Section by section, it moves through continuity at a point and on a domain, the algebra of continuous functions (sums, products, quotients, composites), differentiability and its relationship to continuity, the chain rule for composite functions, implicit differentiation, derivatives of the inverse trigonometric functions, exponential and logarithmic functions and their defining properties, logarithmic differentiation for functions of the form $[u(x)]^{v(x)}$, derivatives of functions given in parametric form, and finally second order derivatives. The guided map below groups these into six teaching blocks used in the core ideas section that follows: continuity at a point; building new continuous functions from old ones; differentiability and its link to continuity; the chain rule together with inverse trigonometric, exponential and logarithmic derivatives; implicit and parametric differentiation; and second order derivatives. Each block ends where the next needs it — you cannot differentiate $\sin^{-1}x$ without the chain rule, and you cannot do logarithmic differentiation without $\dfrac{d}{dx}\log x = \dfrac{1}{x}$ already in hand. Read it in order the first time through; use the cheatsheet at the end for revision.

Continuity at a point — the three-way match

The formal definition sharpens the pen-and-paper picture from the overview. A real function $f$ defined on a subset of $\mathbb{R}$ is continuous at a point $c$ in its domain if the limit of $f$ as $x$ approaches $c$ exists and equals the function's value there.

where: this single equation silently packs in three separate conditions — $f(c)$ must exist (the function is defined at $c$), $\lim_{x\to c} f(x)$ must exist (which itself requires the left hand limit $\lim_{x \to c^-} f(x)$ and right hand limit $\lim_{x \to c^+} f(x)$ to exist and be equal), and finally that common limit value must coincide with $f(c)$. If even one of the three fails, $f$ is discontinuous at $c$, and $c$ is called a point of discontinuity. A function is simply called continuous if it is continuous at every point of its domain; at the endpoints of a closed interval $[a,b]$, continuity means the one-sided limit from inside the interval matches the endpoint value, since the limit from outside is not defined. The step function is the cleanest illustration of failure. Consider $f(x) = 1$ for $x \le 0$ and $f(x) = 2$ for $x > 0$. Near $x=0$, points just to the left give function value $1$ while points just to the right give value $2$; the left hand limit is $1$, the right hand limit is $2$, and since these disagree the limit itself does not exist at $x = 0$, so $f$ is discontinuous there — you would have to lift your pen to draw it. A subtler failure occurs when the one-sided limits do agree with each other but not with $f(c)$, as in $f(x) = x+3$ for $x \ne 0$ and $f(0) = 1$: here $\lim_{x\to0} f(x) = 3 \ne f(0) = 1$, so continuity fails even though there is no jump in the graph's shape, only a single displaced point.

Building new continuous functions from old ones

Checking the definition from scratch at every point is tedious, so the chapter proves an algebra of continuous functions that lets you certify huge classes of functions at once. If $f$ and $g$ are both continuous at $c$, then $f+g$, $f-g$ and $f \cdot g$ are continuous at $c$, and the quotient $f/g$ is continuous at $c$ provided $g(c) \ne 0$.

where: each equality is really the claim that the limit of the combined function equals its value at $c$ — exactly the definition of continuity applied to the new function, and it follows immediately from the corresponding algebra-of-limits rules you already know. The full proof of this result is worked out in the first derivation below. This single theorem, combined with the fact that constant functions, the identity function $f(x)=x$, and $\sin x$ and $\cos x$ are all continuous everywhere, is enough to certify that every polynomial function is continuous (built from sums and products of $x$ and constants) and every rational function $p(x)/q(x)$ is continuous wherever $q(x) \ne 0$. It also handles $\tan x = \sin x/\cos x$, continuous wherever $\cos x \ne 0$. A second theorem covers composition: if $g$ is continuous at $c$ and $f$ is continuous at $g(c)$, then the composite $f \circ g$ is continuous at $c$. This is why functions like $\sin(x^2)$ and $|1 - x + |x||$ are continuous — each is built by feeding one continuous function into another. The one function that resists this good behaviour at isolated points is the greatest integer function $f(x) = [x]$: at every non-integer point the function is locally constant and continuous, but at each integer $n$ the left hand limit is $n-1$ while the right hand limit is $n$, so $[x]$ is discontinuous at every integer.

Differentiability and the derivative revisited

Recall the derivative of $f$ at a point $c$ from Class XI, defined as a limit of the difference quotient — the slope of the secant line as the second point slides in toward $c$.

where: $h$ is the small increment added to $c$; the quotient is the slope of the chord joining $(c, f(c))$ and $(c+h, f(c+h))$, and the limit, when it exists, is the slope of the tangent at $c$. A function is differentiable at $c$ if this limit exists — equivalently, if the left hand derivative $\lim_{h\to 0^-}\frac{f(c+h)-f(c)}{h}$ and the right hand derivative $\lim_{h\to 0^+}\frac{f(c+h)-f(c)}{h}$ both exist and are equal. A function is differentiable on an interval if it is differentiable at every point of that interval, with one-sided derivatives used at the endpoints, exactly as with continuity. The relationship between the two ideas is one-directional, and it is the single most important fact in this chapter: every differentiable function is continuous, but a continuous function need not be differentiable. The classic witness is $f(x) = |x|$: it is continuous at $x=0$ (Example 3 of the NCERT text confirms this directly), yet its left hand derivative there is $-1$ and its right hand derivative is $+1$ — the graph has a sharp corner, so no single tangent line exists, and $f$ fails to be differentiable at $0$ even though you never need to lift your pen to draw it.

The chain rule, and inverse trigonometric, exponential and logarithmic derivatives

Composite functions such as $f(x) = (2x+1)^3$ or $\sin(x^2)$ cannot be differentiated by the basic power, sum or product rules alone. The chain rule handles any function $f = v \circ u$, meaning $f(x) = v(u(x))$, by treating it as a two-stage machine: an inner function $t = u(x)$ feeding into an outer function $v(t)$.

where: $\dfrac{dv}{dt}$ is the derivative of the outer function evaluated at the inner value $t$, and $\dfrac{dt}{dx}$ is the derivative of the inner function; multiplying the two "cancels" the intermediate variable $t$ the same way $\dfrac{a}{b}\cdot\dfrac{b}{c} = \dfrac{a}{c}$ does formally, though the real justification is a limit argument, not algebra. The rule extends to three or more nested functions by repeating the multiplication one layer at a time. The chain rule, combined with implicit differentiation (differentiating a defining relation like $x = \sin y$ directly), produces the three standard inverse trigonometric derivatives, each valid on the stated domain because that is exactly where the corresponding one-to-one branch is defined.

where: the domain restrictions matter — at $x = \pm 1$, $\cos y = 0$ in the underlying derivation and the derivative blows up, which is exactly why $\sin^{-1}x$ and $\cos^{-1}x$ are open-interval results while $\tan^{-1}x$, whose denominator $1+x^2$ never vanishes, holds for every real $x$. The full derivation of the $\sin^{-1}x$ formula is carried out below as a worked derivation. A genuinely new class of functions enters with the exponential function $y = b^x$ for a base $b > 1$, whose most important member is the natural exponential function $y = e^x$, built from the base $e = 1 + \frac{1}{1!} + \frac{1}{2!} + \cdots$, a number strictly between $2$ and $3$. Its inverse is the natural logarithm, written $\log x$ in this chapter (meaning $\log_e x$, sometimes written $\ln x$), defined for $x > 0$ by $\log_b a = x$ if $b^x = a$. Both functions are continuous on their domains, and their derivatives are strikingly simple.

where: the first identity is the defining property that makes $e^x$ indispensable throughout calculus — it is its own derivative, so it never "wears out" under repeated differentiation, unlike any power of $x$. Two algebraic identities of logarithms are used constantly alongside these derivatives: $\log_b(pq) = \log_b p + \log_b q$ and the change of base rule $\log_a p = \dfrac{\log_b p}{\log_b a}$, both of which follow from writing the equal quantities as powers of $b$ and comparing exponents.

Implicit and parametric differentiation

Not every relationship between $x$ and $y$ can be solved for $y$ explicitly — think of $x + \sin(xy) - y = 0$. Even so, $y$ may still depend on $x$, and you can find $\dfrac{dy}{dx}$ by differentiating both sides of the relation directly with respect to $x$, applying the chain rule to every term that contains $y$ (since $y$ is itself a function of $x$, differentiating $y^2$ with respect to $x$ gives $2y\dfrac{dy}{dx}$, not just $2y$). This is implicit differentiation, and it works on any relation $F(x,y) = c$ without ever isolating $y$. A separate situation arises when both $x$ and $y$ are expressed through a third variable, a parameter $t$: $x = f(t)$, $y = g(t)$. The chain rule again supplies the link, since $y$ depends on $x$ only through $t$.

where: $\dfrac{dy}{dt} = g'(t)$ and $\dfrac{dx}{dt} = f'(t)$ are each ordinary derivatives with respect to the single parameter $t$, computed independently, and the ratio gives $\dfrac{dy}{dx}$ purely in terms of $t$ — you are not required (and often not able) to eliminate $t$ and rewrite $y$ as an explicit function of $x$ first. A circle traced as $x = a\cos\theta$, $y = a\sin\theta$ gives $\dfrac{dy}{dx} = \dfrac{a\cos\theta}{-a\sin\theta} = -\cot\theta$, matching the geometric fact that the tangent to a circle is perpendicular to the radius.

Second order derivatives

If $y = f(x)$ is differentiable, the derivative $\dfrac{dy}{dx} = f'(x)$ is itself a new function of $x$. If that function is differentiable, differentiating it once more with respect to $x$ produces the second order derivative, written $\dfrac{d^2y}{dx^2}$ or $f''(x)$ or $y_2$.

where: the notation $\dfrac{d^2y}{dx^2}$ is a single symbol, not a literal square of $\dfrac{d}{dx}$ divided by a square of $dx$ — it records "differentiate twice with respect to $x$." When $y$ is given implicitly or parametrically, finding $\dfrac{d^2y}{dx^2}$ means differentiating the expression for $\dfrac{dy}{dx}$ a second time with respect to $x$, which usually still contains $y$ or the parameter and so needs the chain rule (or implicit differentiation) all over again. Second order derivatives are the tool this chapter hands directly to the next one, Application of Derivatives, where the sign of $f''(x)$ decides concavity and distinguishes maxima from minima.

Board level - continuity of a piecewise linear function at the join point (NCERT Example 10, extended)

Discuss the continuity of $f(x) = x+2$ for $x \le 1$ and $f(x) = x-2$ for $x > 1$, and state all points of discontinuity, if any.

Step 1. Check continuity away from the join. For any $c < 1$, $f$ coincides with $x+2$ in a neighbourhood of $c$, so $\lim_{x\to c}f(x) = c+2 = f(c)$, and $f$ is continuous there. For any $c > 1$, $f$ coincides with $x-2$ nearby, so $\lim_{x\to c}f(x) = c-2 = f(c)$, again continuous. The only point that needs care is $c = 1$, where the formula switches.

Step 2. Compute the left hand limit at $x=1$, using the branch $x+2$ that applies for $x \le 1$. $\lim_{x\to 1^-}f(x) = \lim_{x\to 1^-}(x+2) = 1+2 = 3$.

Step 3. Compute the right hand limit at $x=1$, using the branch $x-2$ that applies for $x>1$. $\lim_{x\to 1^+}f(x) = \lim_{x\to 1^+}(x-2) = 1-2 = -1$.

Step 4. Compare. Since $3 \ne -1$, the left and right hand limits at $x=1$ disagree, so $\lim_{x\to 1}f(x)$ does not exist and $f$ is discontinuous at $x=1$ regardless of what value $f(1)$ actually takes (here $f(1) = 1+2 = 3$, matching the left branch, so the jump is purely a right-side mismatch). Answer: $f$ is continuous at every real number except $x = 1$, where it has a jump discontinuity from $3$ down to $-1$.

Board level - continuity versus differentiability of the shifted modulus function (NCERT Exercise 5.2, Q9, extended)

Show that $f(x) = |x-1|$ is continuous for all real $x$ but not differentiable at $x=1$.

Step 1. Establish continuity. Write $f(x) = |x-1|$ as the composite $h(g(x))$ where $g(x) = x-1$ (a polynomial, continuous everywhere) and $h(t) = |t|$ (continuous everywhere, as shown for $|x|$ using left and right hand limits both equal to $0$ at $t=0$). By the composition theorem for continuous functions, $f = h\circ g$ is continuous at every real number, including $x=1$.

Step 2. Set up the derivative check at $x=1$ using the definition, with $f(1) = |1-1| = 0$. $\dfrac{f(1+h)-f(1)}{h} = \dfrac{|h|-0}{h} = \dfrac{|h|}{h}$.

Step 3. Compute the left hand derivative, where $h \to 0^-$ so $h < 0$ and $|h| = -h$. $\lim_{h\to 0^-}\dfrac{|h|}{h} = \lim_{h\to 0^-}\dfrac{-h}{h} = -1$.

Step 4. Compute the right hand derivative, where $h \to 0^+$ so $h>0$ and $|h|=h$. $\lim_{h\to 0^+}\dfrac{|h|}{h} = \lim_{h\to 0^+}\dfrac{h}{h} = 1$. Since $-1 \ne 1$, the two one-sided derivatives disagree, so $f'(1)$ does not exist. Answer: $f(x)=|x-1|$ is continuous everywhere but not differentiable at $x=1$, where the graph has a sharp corner exactly like $|x|$ at the origin, shifted one unit to the right.

Competitive level (JEE / NEET) - logarithmic differentiation of a sum of three variable-exponent terms (NCERT Example 30, extended)

If $y^x + x^y + x^x = a^b$ for constants $a, b$, find $\dfrac{dy}{dx}$.

Step 1. Split into three pieces and differentiate the constant. Let $u = y^x$, $v = x^y$, $w = x^x$, so $u+v+w = a^b$, a constant. Differentiating both sides with respect to $x$ gives $\dfrac{du}{dx}+\dfrac{dv}{dx}+\dfrac{dw}{dx} = 0$, since the derivative of a constant is zero. Each term now needs logarithmic differentiation separately, because each has a variable in both base and exponent (or, for $w$, a variable exponent on a variable base).

Step 2. Differentiate $u = y^x$. Taking logs, $\log u = x\log y$. Differentiating with the product rule on the right and the chain rule on the left (since $u$ depends on $x$ through $y$ too): $\dfrac{1}{u}\dfrac{du}{dx} = x\cdot\dfrac{1}{y}\dfrac{dy}{dx} + \log y$, so $\dfrac{du}{dx} = y^x\left(\dfrac{x}{y}\dfrac{dy}{dx} + \log y\right)$.

Step 3. Differentiate $v = x^y$. Taking logs, $\log v = y\log x$. Differentiating: $\dfrac{1}{v}\dfrac{dv}{dx} = y\cdot\dfrac{1}{x} + \dfrac{dy}{dx}\log x$, so $\dfrac{dv}{dx} = x^y\left(\dfrac{y}{x} + \dfrac{dy}{dx}\log x\right)$. Then differentiate $w = x^x$ the same way: $\log w = x\log x$, so $\dfrac{1}{w}\dfrac{dw}{dx} = x\cdot\dfrac{1}{x} + \log x = 1+\log x$, giving $\dfrac{dw}{dx} = x^x(1+\log x)$.

Step 4. Substitute all three derivatives into $\dfrac{du}{dx}+\dfrac{dv}{dx}+\dfrac{dw}{dx}=0$ and collect every term containing $\dfrac{dy}{dx}$ on one side. $y^x\left(\dfrac{x}{y}\dfrac{dy}{dx}+\log y\right) + x^y\left(\dfrac{y}{x}+\dfrac{dy}{dx}\log x\right) + x^x(1+\log x) = 0$. Grouping the $\dfrac{dy}{dx}$ terms: $\left(x\,y^{x-1} + x^y\log x\right)\dfrac{dy}{dx} = -\left[y^x\log y + y\,x^{y-1} + x^x(1+\log x)\right]$.

Step 5. Isolate $\dfrac{dy}{dx}$. Answer:

Timing note: aim for under four minutes. The whole problem is three parallel applications of the same logarithmic-differentiation template, so the speed gain comes from recognising the pattern instantly ("variable to a variable power, so log first") rather than re-deriving it each time — write $\log u = x\log y$, $\log v = y\log x$, $\log w = x\log x$ in one line before differentiating anything, then grind through the product rule on each. The most common slip under time pressure is forgetting that $y$ is a function of $x$ when differentiating $x\log y$, dropping the $\frac{dy}{dx}$ that the chain rule demands.

Competitive level (JEE / NEET) - second order derivative of a parametric curve (NCERT Miscellaneous Exercise, Q17, extended)

If $x = a(\cos t + t\sin t)$ and $y = a(\sin t - t\cos t)$, find $\dfrac{d^2y}{dx^2}$.

Step 1. Differentiate $x$ and $y$ with respect to the parameter $t$ separately, using the product rule on the $t\sin t$ and $t\cos t$ terms. $\dfrac{dx}{dt} = a\left(-\sin t + \sin t + t\cos t\right) = a\,t\cos t$, since $-\sin t$ and $+\sin t$ cancel. Similarly, $\dfrac{dy}{dt} = a\left(\cos t - \cos t + t\sin t\right) = a\,t\sin t$, since $\cos t$ and $-\cos t$ cancel.

Step 2. Form the first derivative using $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$. $\dfrac{dy}{dx} = \dfrac{a\,t\sin t}{a\,t\cos t} = \dfrac{\sin t}{\cos t} = \tan t$, valid for $t \ne 0$ (where $dx/dt = 0$) and $\cos t \ne 0$. Notice the parameter $a$ and the factor $t$ both cancel completely — a strong hint that the algebra was done correctly.

Step 3. Differentiate $\dfrac{dy}{dx} = \tan t$ with respect to $x$, which requires the chain rule since $\tan t$ is a function of $t$, not directly of $x$: $\dfrac{d}{dx}(\tan t) = \dfrac{d}{dt}(\tan t)\cdot\dfrac{dt}{dx} = \sec^2 t \cdot \dfrac{dt}{dx}$.

Step 4. Replace $\dfrac{dt}{dx}$ using $\dfrac{dt}{dx} = \dfrac{1}{dx/dt} = \dfrac{1}{a\,t\cos t}$, from Step 1. Substituting, $\dfrac{d^2y}{dx^2} = \sec^2 t \cdot \dfrac{1}{a\,t\cos t} = \dfrac{1}{a\,t\cos^3 t}$, using $\sec^2 t \cdot \dfrac{1}{\cos t} = \dfrac{1}{\cos^2 t}\cdot\dfrac{1}{\cos t} = \dfrac{1}{\cos^3 t}$. Answer:

Timing note: aim for under three minutes once the pattern is automatic. The parametric second-derivative recipe never changes: find $dy/dt$ and $dx/dt$, divide for $dy/dx$, then differentiate that expression with respect to $t$ and divide by $dx/dt$ again — never attempt to eliminate $t$ and differentiate an explicit $y(x)$ unless the algebra is unusually friendly. The biggest time sink is forgetting the second division by $dx/dt$ in Step 4 and submitting $\sec^2 t$ as if it were already $\dfrac{d^2y}{dx^2}$, which throws away the entire chain-rule step.