Chapter 5 · Continuity and Differentiability

5.2 Differentiability, the Chain Rule and Implicit Functions

A function is differentiable at $c$ when its left-hand and right-hand derivatives agree; differentiability forces continuity, but not conversely. The chain rule, implicit differentiation and the derivatives of $\sin^{-1}x$, $\tan^{-1}x$, … let us differentiate almost anything.

A cyclist rides at a steady speed, then brakes hard at a junction. Her distance from home is a continuous function of time, and so is her speed — until the instant the brakes bite, when the speed starts to drop at a sudden new rate. On a graph of distance against time that instant may look harmless, but on the graph of speed it is a corner. Corners are where the derivative — the slope of the graph — fails to exist, even though nothing breaks. The first part of this section tells you exactly how to detect them, and proves the one relation between the ideas that you must never get backwards: a differentiable function is always continuous, but a continuous function need not be differentiable.

The rest of the section extends your Class 11 rules (sum, product, quotient) to three new situations. Most functions in physics and engineering are layered — the height of a piston is $\sin(\omega t)$, the voltage of a circuit is $\sqrt{1+v^2}$ — and the chain rule says how rates multiply through the layers. Some curves are given only by an equation linking $x$ and $y$, like the circle $x^2+y^2=25$; implicit differentiation finds their slopes without solving for $y$. And the inverse trigonometric functions of Chapter 2 turn out to have surprisingly simple algebraic derivatives, which you will use constantly when integrating in Chapter 7.

Differentiability and Its Link with Continuity

From Class 11 Mathematics, Ch. 12 — Limits and Derivatives: the derivative of $f$ at $c$ is $f'(c)=\lim\limits_{h\to0}\dfrac{f(c+h)-f(c)}{h}$, the limit of slopes of chords through $(c,f(c))$. Like any limit, it can be split into two one-sided limits.

📘 Definition 5.3 — Left-hand and right-hand derivatives; differentiability

Let $f$ be a real function defined on an open interval containing $c$. The left-hand derivative and right-hand derivative of $f$ at $c$ are $$Lf'(c)=\lim_{h\to0^-}\dfrac{f(c+h)-f(c)}{h},\qquad Rf'(c)=\lim_{h\to0^+}\dfrac{f(c+h)-f(c)}{h},$$ when these limits exist (as finite numbers). $f$ is differentiable at $c$ if both exist and are equal; their common value is the derivative $f'(c)$.

$f$ is differentiable on an open interval $(a,b)$ if it is differentiable at every point of $(a,b)$; on $[a,b]$ if, in addition, $Rf'(a)$ and $Lf'(b)$ exist. A differentiable function is one that is differentiable at every point of its domain.

✏️ Example 1 — One-sided derivatives at a corner
(a) Show that $f(x)=|x|$ is not differentiable at $x=0$. (b) Is $g(x)=|x-2|$ differentiable at $x=2$? (c) Show that $k(x)=x|x|$ is differentiable at $x=0$, and find $k'(0)$.
(a) $\dfrac{f(0+h)-f(0)}{h}=\dfrac{|h|}{h}$, which is $-1$ for $h\lt0$ and $1$ for $h\gt0$. So $Lf'(0)=-1\ne1=Rf'(0)$: not differentiable at $0$ — even though $|x|$ is continuous there (Section 5.1, Example 2).
(b) $\dfrac{g(2+h)-g(2)}{h}=\dfrac{|h|}{h}$ again: $Lg'(2)=-1$, $Rg'(2)=1$. Not differentiable at $2$ — the same corner, shifted.
(c) $\dfrac{k(0+h)-k(0)}{h}=\dfrac{h|h|}{h}=|h|\to0$ from both sides. So $Lk'(0)=Rk'(0)=0$: differentiable, $k'(0)=0$. Multiplying $|x|$ by $x$ flattens the corner.
📐 Theorem 5.4 — Differentiability implies continuity

If $f$ is differentiable at $c$, then $f$ is continuous at $c$.

Proof. For $h\ne0$, $$f(c+h)-f(c)=\dfrac{f(c+h)-f(c)}{h}\cdot h.$$ As $h\to0$ the first factor tends to $f'(c)$ (a finite number) and the second to $0$, so by the product rule for limits $f(c+h)-f(c)\to f'(c)\cdot0=0$. That is, $\lim\limits_{h\to0}f(c+h)=f(c)$, which says $f$ is continuous at $c$. $\blacksquare$

The converse is false: $|x|$ is continuous but not differentiable at $0$. Equivalently (the contrapositive), a function that is not continuous at $c$ cannot be differentiable at $c$.

✏️ Example 2 — Two ways to fail
(a) Show that $f(x)=[x]$ is not differentiable at $x=1$. (b) Show that $g(x)=x^{1/3}$ is continuous but not differentiable at $x=0$, and describe its graph there.
(a) $[x]$ is not continuous at $1$ (Section 5.1), so by Theorem 5.4 it is not differentiable there. Directly: for $-1\lt h\lt0$, $\dfrac{[1+h]-[1]}{h}=\dfrac{0-1}{h}=-\dfrac1h\to+\infty$, so $Lf'(1)$ does not exist.
(b) continuous $x^{1/3}$ is the real cube root; $\lim\limits_{x\to0}x^{1/3}=0=g(0)$.
(b) not differentiable $\dfrac{g(h)-g(0)}{h}=\dfrac{h^{1/3}}{h}=\dfrac{1}{h^{2/3}}\to+\infty$ as $h\to0$ (from either side). The limit is not a finite number, so $g'(0)$ does not exist.
Picture The chords through the origin get steeper and steeper: the graph has a vertical tangent at $0$. A vertical line has no slope, so there is no derivative — a third way to fail, besides a break and a corner.
✏️ Example 3 — Making a join smooth
Find $a$ and $b$ so that $f(x)=\begin{cases}x^2,&x\le1\\ ax+b,&x\gt1\end{cases}$ is differentiable at $x=1$.
Continuity first Differentiable forces continuous (Theorem 5.4): LHL $=f(1)=1$ and RHL $=a+b$, so $a+b=1$.
Left derivative $Lf'(1)=\lim\limits_{h\to0^-}\dfrac{(1+h)^2-1}{h}=\lim\limits_{h\to0^-}(2+h)=2$.
Right derivative Using $a+b=1$: $Rf'(1)=\lim\limits_{h\to0^+}\dfrac{a(1+h)+b-1}{h}=\lim\limits_{h\to0^+}\dfrac{ah}{h}=a$.
Answer $a=2$, $b=-1$. The line $y=2x-1$ is exactly the tangent to $y=x^2$ at $(1,1)$, so the two pieces join without a corner.
Why continuity first? If you skip it you may get the right $a$ but a wrong $b$: with $a+b\ne1$ the quotient $\tfrac{a(1+h)+b-1}{h}$ blows up, and $Rf'(1)$ does not exist at all.
Interactive 1 — Zoom In: Smooth Points Straighten, Corners Do Not

The window is centred on the point $(c,f(c))$ and keeps equal scales on both axes. Play zooms in by a factor of $1000$ and back out. At a point where $f$ is differentiable, the curve straightens into a line — the tangent. At a corner it stays a corner however far you zoom. The red and blue chords give the left and right difference quotients for $h$ equal to half the window.

half-width $h$—
left quotient $\tfrac{f(c)-f(c-h)}{h}$—
right quotient $\tfrac{f(c+h)-f(c)}{h}$—
verdict at $c$—

The Chain Rule

Think of three meshed gears. If gear $A$ turns $3$ times for every turn of gear $B$, and $B$ turns $2$ times for every turn of gear $C$, then $A$ turns $3\times2=6$ times for every turn of $C$. Rates of change through a chain of dependencies multiply. For a layered function $y=v(t)$ with $t=u(x)$, a small change $\Delta x$ causes a change $\Delta t$, which causes a change $\Delta y$, and (when $\Delta t\ne0$) $$\dfrac{\Delta y}{\Delta x}=\dfrac{\Delta y}{\Delta t}\cdot\dfrac{\Delta t}{\Delta x}.$$ Letting $\Delta x\to0$ suggests the rule below. (A complete proof also handles the case where $\Delta t$ is $0$ for some small $\Delta x$; we accept the result.)

📐 Theorem 5.5 — The chain rule

Let $f=v\circ u$, that is $f(x)=v(u(x))$. If $u$ is differentiable at $x$ and $v$ is differentiable at $u(x)$, then $f$ is differentiable at $x$ and $$f'(x)=v'(u(x))\cdot u'(x).$$ In Leibniz notation, with $t=u(x)$ and $y=v(t)$: $$\dfrac{dy}{dx}=\dfrac{dy}{dt}\cdot\dfrac{dt}{dx}.$$ For three layers $y=w(s)$, $s=v(t)$, $t=u(x)$: $\dfrac{dy}{dx}=\dfrac{dy}{ds}\cdot\dfrac{ds}{dt}\cdot\dfrac{dt}{dx}$.

In words: differentiate the outside layer, leaving the inside alone, then multiply by the derivative of the inside. So $\dfrac{d}{dx}\sin(\text{stuff})=\cos(\text{stuff})\cdot\dfrac{d}{dx}(\text{stuff})$ and $\dfrac{d}{dx}(\text{stuff})^n=n(\text{stuff})^{n-1}\cdot\dfrac{d}{dx}(\text{stuff})$.

✏️ Example 4 — Peeling the layers
Differentiate: (a) $\sin(x^2)$   (b) $(2x+3)^5$   (c) $\tan\sqrt x$ ($x\gt0$)   (d) $\sin(\cos(x^2))$
(a) Outside $\sin t$, inside $t=x^2$: $\dfrac{d}{dx}\sin(x^2)=\cos(x^2)\cdot2x=2x\cos(x^2)$.
(b) Outside $t^5$, inside $t=2x+3$: $5(2x+3)^4\cdot2=10(2x+3)^4$. (Expanding the fifth power first would work, but takes ten times as long.)
(c) Outside $\tan t$, inside $t=\sqrt x$ with $\dfrac{dt}{dx}=\dfrac{1}{2\sqrt x}$: $\dfrac{d}{dx}\tan\sqrt x=\sec^2\!\sqrt x\cdot\dfrac{1}{2\sqrt x}=\dfrac{\sec^2\sqrt x}{2\sqrt x}$.
(d) Three layers: $y=\sin s$, $s=\cos t$, $t=x^2$. $\dfrac{dy}{dx}=\cos s\cdot(-\sin t)\cdot2x=-2x\sin(x^2)\cos(\cos(x^2))$.
Check (a) at $x=1$: $2\cos1\approx1.0806$; the difference quotient $\dfrac{\sin(1.0001^2)-\sin(0.9999^2)}{0.0002}\approx1.0806$ ✓.
✏️ Example 5 — The chain rule with the product and quotient rules
Find $\dfrac{dy}{dx}$: (a) $y=x\sqrt{1+x^2}$   (b) $y=\cos^3(2x)$   (c) $y=(x^2+1)^3(2x-1)^2$, giving the answer in factorised form.
(a) Product rule, then the chain rule for $\sqrt{1+x^2}$, whose derivative is $\dfrac{2x}{2\sqrt{1+x^2}}=\dfrac{x}{\sqrt{1+x^2}}$: $$\dfrac{dy}{dx}=\sqrt{1+x^2}+\dfrac{x^2}{\sqrt{1+x^2}}=\dfrac{1+2x^2}{\sqrt{1+x^2}}.$$
(b) $\cos^3(2x)$ means $(\cos2x)^3$: three layers, cube $\to$ cosine $\to$ $2x$. $\dfrac{dy}{dx}=3\cos^2(2x)\cdot(-\sin2x)\cdot2=-6\cos^2(2x)\sin(2x)$.
(c) $\dfrac{dy}{dx}=3(x^2+1)^2(2x)\cdot(2x-1)^2+(x^2+1)^3\cdot2(2x-1)(2)$. Take out the common factor $2(x^2+1)^2(2x-1)$: $$\dfrac{dy}{dx}=2(x^2+1)^2(2x-1)\big[3x(2x-1)+2(x^2+1)\big]=2(x^2+1)^2(2x-1)(8x^2-3x+2).$$
Check (c) at $x=1$: $2\cdot4\cdot1\cdot7=56$. Directly, $y(1.0001)-y(0.9999)\approx0.0112$, divided by $0.0002$ gives $56$ ✓.

Implicit Differentiation

An equation such as $y=x^3-2x$ gives $y$ explicitly as a function of $x$. An equation such as $x^2+y^2=25$ links $x$ and $y$ but does not single out $y$: it is satisfied by $y=\sqrt{25-x^2}$ and by $y=-\sqrt{25-x^2}$. Near most points of such a curve, $y$ is still a differentiable function of $x$ — it is defined implicitly by the equation — and we can find $\dfrac{dy}{dx}$ without solving for $y$ at all.

🧭 Problem-solving strategy — implicit differentiation
  1. Differentiate both sides of the equation with respect to $x$, treating $y$ as an unknown function of $x$.
  2. Every time you differentiate an expression in $y$, the chain rule adds a factor $\dfrac{dy}{dx}$: $\dfrac{d}{dx}(y^2)=2y\dfrac{dy}{dx}$, $\dfrac{d}{dx}(\sin y)=\cos y\dfrac{dy}{dx}$. A product such as $xy$ needs the product rule: $\dfrac{d}{dx}(xy)=y+x\dfrac{dy}{dx}$.
  3. Collect the terms containing $\dfrac{dy}{dx}$ on one side, factor it out, and divide.
  4. The answer usually involves both $x$ and $y$. To find the slope at a point, substitute both coordinates — and check the point lies on the curve.
✏️ Example 6 — Slopes of implicit curves
Find $\dfrac{dy}{dx}$: (a) $x^2+y^2=25$, and its value at $(3,4)$   (b) $x^3+y^3=6xy$, and its value at $(3,3)$   (c) $xy+y^2=\tan x+y$
(a) $2x+2y\dfrac{dy}{dx}=0$, so $\dfrac{dy}{dx}=-\dfrac xy$ ($y\ne0$). At $(3,4)$, which is on the circle since $9+16=25$: slope $-\dfrac34$. Check with the explicit upper half $y=\sqrt{25-x^2}$: $y'=\dfrac{-x}{\sqrt{25-x^2}}=-\dfrac34$ at $x=3$ ✓.
(b) $3x^2+3y^2\dfrac{dy}{dx}=6y+6x\dfrac{dy}{dx}$. Dividing by $3$ and collecting: $(y^2-2x)\dfrac{dy}{dx}=2y-x^2$, so $\dfrac{dy}{dx}=\dfrac{2y-x^2}{y^2-2x}$ (where $y^2\ne2x$). $(3,3)$ is on the curve ($27+27=54=6\cdot9$), and the slope there is $\dfrac{6-9}{9-6}=-1$.
(c) $y+x\dfrac{dy}{dx}+2y\dfrac{dy}{dx}=\sec^2x+\dfrac{dy}{dx}$, so $(x+2y-1)\dfrac{dy}{dx}=\sec^2x-y$ and $\dfrac{dy}{dx}=\dfrac{\sec^2x-y}{x+2y-1}$.
Notice In (b) the slope formula fails where $y^2=2x$: there the tangent is vertical. Interactive 2 shows such points on the curve.
Interactive 2 — Tangents to Implicit Curves

Choose a curve and drag the point along it (or press Play). The slope of the tangent line is computed from the implicit-differentiation formula using only the point's coordinates $(x,y)$ — no solving for $y$. Find the points where the tangent is horizontal (numerator $0$) and vertical (denominator $0$).

point $(x,y)$—
formula for $\tfrac{dy}{dx}$—
slope at the point—
on the curve? (LHS − RHS)—

Derivatives of Inverse Trigonometric Functions

Implicit differentiation gives the derivatives of the inverse trigonometric functions almost at once. Recall from Chapter 2 — Inverse Trigonometric Functions that $y=\sin^{-1}x$ means $\sin y=x$ with $y$ in the principal value branch $\left[-\tfrac\pi2,\tfrac\pi2\right]$. (Remember $\sin^{-1}x\ne\dfrac1{\sin x}$: the $-1$ denotes the inverse function, not a reciprocal.)

📐 Theorem 5.6 — Derivatives of the inverse trigonometric functions
$f(x)$$f'(x)$valid for
$\sin^{-1}x$$\dfrac{1}{\sqrt{1-x^2}}$$-1\lt x\lt1$
$\cos^{-1}x$$-\dfrac{1}{\sqrt{1-x^2}}$$-1\lt x\lt1$
$\tan^{-1}x$$\dfrac{1}{1+x^2}$all real $x$
$\cot^{-1}x$$-\dfrac{1}{1+x^2}$all real $x$
$\sec^{-1}x$$\dfrac{1}{|x|\sqrt{x^2-1}}$$|x|\gt1$
$\operatorname{cosec}^{-1}x$$-\dfrac{1}{|x|\sqrt{x^2-1}}$$|x|\gt1$

Proof for $\sin^{-1}x$. Let $y=\sin^{-1}x$ with $-1\lt x\lt1$, so $\sin y=x$ and $-\tfrac\pi2\lt y\lt\tfrac\pi2$. Differentiating, $\cos y\,\dfrac{dy}{dx}=1$. On $\left(-\tfrac\pi2,\tfrac\pi2\right)$ the cosine is positive, so $\cos y=+\sqrt{1-\sin^2y}=\sqrt{1-x^2}\ne0$, and $\dfrac{dy}{dx}=\dfrac{1}{\sqrt{1-x^2}}$. (At $x=\pm1$ the tangent to $y=\sin^{-1}x$ is vertical: no derivative.)

Proof for $\tan^{-1}x$. $y=\tan^{-1}x$ means $\tan y=x$ with $-\tfrac\pi2\lt y\lt\tfrac\pi2$. Then $\sec^2y\,\dfrac{dy}{dx}=1$ and $\sec^2y=1+\tan^2y=1+x^2$, so $\dfrac{dy}{dx}=\dfrac{1}{1+x^2}$.

The others follow in the same way (Exercise IV). For $x\gt1$ the $\sec^{-1}$ entry is simply $\dfrac{1}{x\sqrt{x^2-1}}$; the modulus matters only for $x\lt-1$.

✏️ Example 7 — Inverse trigonometric derivatives
Find $\dfrac{dy}{dx}$: (a) $y=\sin^{-1}(2x)$, $|x|\lt\tfrac12$   (b) $y=\tan^{-1}(x^2)$   (c) $y=\sin^{-1}\!\left(2x\sqrt{1-x^2}\right)$, $-\tfrac{1}{\sqrt2}\lt x\lt\tfrac{1}{\sqrt2}$   (d) $y=\tan^{-1}\!\left(\dfrac{3x-x^3}{1-3x^2}\right)$, $-\tfrac{1}{\sqrt3}\lt x\lt\tfrac{1}{\sqrt3}$
(a) Chain rule with inside $2x$: $\dfrac{1}{\sqrt{1-(2x)^2}}\cdot2=\dfrac{2}{\sqrt{1-4x^2}}$.
(b) $\dfrac{1}{1+(x^2)^2}\cdot2x=\dfrac{2x}{1+x^4}$.
(c) simplify first Put $x=\sin\theta$ with $\theta=\sin^{-1}x$. The condition on $x$ gives $-\tfrac\pi4\lt\theta\lt\tfrac\pi4$, and $\sqrt{1-x^2}=\cos\theta$ (positive). So $2x\sqrt{1-x^2}=2\sin\theta\cos\theta=\sin2\theta$ with $-\tfrac\pi2\lt2\theta\lt\tfrac\pi2$ — inside the principal branch. Hence $y=\sin^{-1}(\sin2\theta)=2\theta=2\sin^{-1}x$ and $\dfrac{dy}{dx}=\dfrac{2}{\sqrt{1-x^2}}$.
(d) simplify first Put $x=\tan\theta$, $-\tfrac\pi6\lt\theta\lt\tfrac\pi6$. Then $\dfrac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}=\tan3\theta$ with $-\tfrac\pi2\lt3\theta\lt\tfrac\pi2$, so $y=3\theta=3\tan^{-1}x$ and $\dfrac{dy}{dx}=\dfrac{3}{1+x^2}$.
Check (c) at $x=0.3$: the formula gives $\dfrac{2}{\sqrt{0.91}}\approx2.0966$; a difference quotient of $\sin^{-1}(2x\sqrt{1-x^2})$ with step $10^{-5}$ gives the same ✓. The interval matters: for $x$ between $\tfrac1{\sqrt2}$ and $1$, $2\theta$ leaves the principal branch and the simplification changes.
⚠️ Common mistakes when studying alone

1. Getting the implication backwards. Differentiable $\Rightarrow$ continuous. Continuous does not imply differentiable ($|x|$ at $0$). To show a function is not differentiable, showing it is discontinuous is enough; to show it is differentiable, continuity is not enough.

2. Differentiating the pieces instead of using the definition. For a piecewise function, the derivatives of the two formulas at the join give $Lf'(c)$ and $Rf'(c)$ only if $f$ is continuous at $c$ (and each piece is differentiable there). Check continuity first (Example 3).

3. Forgetting the inside derivative. $\dfrac{d}{dx}\sin(5x)=5\cos5x$, not $\cos5x$; $\dfrac{d}{dx}(1-x^2)^{1/2}=\tfrac12(1-x^2)^{-1/2}\cdot(-2x)$.

4. Changing the inside. $\dfrac{d}{dx}\sin(x^2)$ is $\cos(x^2)\cdot2x$, not $\cos(2x)$: the outer derivative is evaluated at the unchanged inside.

5. Dropping $\dfrac{dy}{dx}$ in implicit work. $\dfrac{d}{dx}(y^3)=3y^2\dfrac{dy}{dx}$, and $\dfrac{d}{dx}(xy)=y+x\dfrac{dy}{dx}$ — not $\dfrac{dy}{dx}$ alone.

6. Simplifying inverse trigonometric expressions without checking the range. $\sin^{-1}(\sin2\theta)=2\theta$ only when $2\theta$ lies in $\left[-\tfrac\pi2,\tfrac\pi2\right]$.

🚀 Beyond the board syllabus (JEE Main · JEE Advanced)

In JEE Main and JEE Advanced 2026; not examined in the CBSE board exam. JEE asks where modulus-type functions fail to be differentiable. A useful rule: if $g$ is differentiable, then $|g(x)|$ is differentiable wherever $g(x)\ne0$; at a zero $c$ of $g$ it is differentiable if and only if $g'(c)=0$. (Because near $c$, $\dfrac{|g(c+h)|}{h}\approx\dfrac{|g'(c)|\,|h|}{h}$, whose one-sided limits are $\pm|g'(c)|$.)

Worked example. Where is $f(x)=|x^2-4|+|x|^3$ not differentiable?

$g(x)=x^2-4$ has zeros $\pm2$ with $g'(\pm2)=\pm4\ne0$: corners at $x=\pm2$. The term $|x|^3$ is $|g|$ for $g(x)=x^3$, whose only zero $0$ has $g'(0)=0$: it is differentiable everywhere, with derivative $3x|x|$. A sum is non-differentiable where exactly one term is; so $f$ fails to be differentiable exactly at $x=-2$ and $x=2$. Check at $2$: $Rf'(2)=4+12=16$ and $Lf'(2)=-4+12=8$.

Exercises

I
Differentiability at a point.
  • 1 Is $f(x)=|x+3|$ differentiable at $x=-3$? Find $Lf'(-3)$ and $Rf'(-3)$.
  • 2 $f(x)=\begin{cases}x+1,&x\le1\\ 3-x,&x\gt1\end{cases}$. Is $f$ continuous at $1$? differentiable at $1$?
  • 3 $f(x)=\begin{cases}x^2,&x\le2\\ 4x-4,&x\gt2\end{cases}$. Show that $f$ is differentiable at $2$ and find $f'(2)$.
  • 4 Is $f(x)=[x]$ differentiable at $x=0.5$? at $x=2$? Give $f'(0.5)$ if it exists.
  • 5 Show that $f(x)=x^{2/3}$ is continuous but not differentiable at $x=0$.
  • 6 Find $a$ and $b$ so that $f(x)=\begin{cases}ax^2+b,&x\le1\\ \dfrac1x,&x\gt1\end{cases}$ is differentiable at $x=1$.
  • 7 True or false? Give a reason. (a) Every continuous function is differentiable. (b) If $f$ is not continuous at $c$, it is not differentiable at $c$. (c) $f(x)=|x|+|x-1|$ is differentiable at $x=\tfrac12$.
1 $\dfrac{f(-3+h)-f(-3)}{h}=\dfrac{|h|}{h}$: $Lf'(-3)=-1$, $Rf'(-3)=1$. Not differentiable.
2 LHL $=f(1)=2$, RHL $=2$: continuous. $Lf'(1)=\lim\limits_{h\to0^-}\dfrac{(2+h)-2}{h}=1$, $Rf'(1)=\lim\limits_{h\to0^+}\dfrac{(2-h)-2}{h}=-1$. Not differentiable (a peak-shaped corner).
3 Continuous: $4=4$. $Lf'(2)=\lim\dfrac{(2+h)^2-4}{h}=\lim(4+h)=4$; $Rf'(2)=\lim\dfrac{4(2+h)-4-4}{h}=4$. Differentiable, $f'(2)=4$.
4 Near $0.5$, $[x]=0$ is constant, so $f'(0.5)=0$. At $2$, $[x]$ is not continuous, so it is not differentiable (Theorem 5.4).
5 $\lim\limits_{x\to0}x^{2/3}=0=f(0)$: continuous. $\dfrac{h^{2/3}}{h}=h^{-1/3}$, which tends to $+\infty$ as $h\to0^+$ and to $-\infty$ as $h\to0^-$. No derivative (a cusp).
6 Continuity: $a+b=1$. With that, $Lf'(1)=\lim\limits_{h\to0^-}\dfrac{a(1+h)^2+b-1}{h}=\lim(2a+ah)=2a$ and $Rf'(1)=\lim\limits_{h\to0^+}\dfrac{\frac{1}{1+h}-1}{h}=\lim\dfrac{-1}{1+h}=-1$. So $2a=-1$: $a=-\tfrac12$, $b=\tfrac32$.
7 (a) False: $|x|$ at $0$. (b) True: it is the contrapositive of Theorem 5.4. (c) True: for $0\lt x\lt1$, $f(x)=x+(1-x)=1$, a constant near $\tfrac12$, so $f'(\tfrac12)=0$.
II
Differentiate using the chain rule.
  • 1 $\sin(3x+2)$
  • 2 $(x^2-5x+1)^7$
  • 3 $\cos\sqrt x$, $x\gt0$
  • 4 $\sec^2(4x)$
  • 5 $\sqrt{x^2-2x+3}$
  • 6 $\sin^3x\cos^3x$
  • 7 $\dfrac{1}{(2x-1)^3}$
  • 8 $\sin(\tan(x^2))$
  • 9 $\dfrac{x}{\sqrt{1-x^2}}$, $|x|\lt1$
  • 10 A weather balloon. The radius of a spherical balloon increases at $0.5\ \text{cm/s}$. Using $V=\tfrac43\pi r^3$ and the chain rule, find how fast its volume is increasing when $r=10\ \text{cm}$.
1 $3\cos(3x+2)$
2 $7(x^2-5x+1)^6(2x-5)$
3 $-\sin\sqrt x\cdot\dfrac{1}{2\sqrt x}=-\dfrac{\sin\sqrt x}{2\sqrt x}$
4 $2\sec(4x)\cdot\sec(4x)\tan(4x)\cdot4=8\sec^2(4x)\tan(4x)$
5 $\dfrac{2x-2}{2\sqrt{x^2-2x+3}}=\dfrac{x-1}{\sqrt{x^2-2x+3}}$ (the square root is always defined, since $x^2-2x+3=(x-1)^2+2\gt0$)
6 $3\sin^2x\cos x\cdot\cos^3x+\sin^3x\cdot3\cos^2x(-\sin x)=3\sin^2x\cos^2x(\cos^2x-\sin^2x)=3\sin^2x\cos^2x\cos2x$
7 $(2x-1)^{-3}$ gives $-3(2x-1)^{-4}\cdot2=-\dfrac{6}{(2x-1)^4}$
8 $\cos(\tan x^2)\cdot\sec^2(x^2)\cdot2x$
9 $\dfrac{\sqrt{1-x^2}-x\cdot\frac{-x}{\sqrt{1-x^2}}}{1-x^2}=\dfrac{(1-x^2)+x^2}{(1-x^2)^{3/2}}=\dfrac{1}{(1-x^2)^{3/2}}$
10 $\dfrac{dV}{dt}=\dfrac{dV}{dr}\cdot\dfrac{dr}{dt}=4\pi r^2\cdot0.5=4\pi(100)(0.5)=200\pi\approx628\ \text{cm}^3\text{/s}$.
III
Implicit differentiation. Find $\dfrac{dy}{dx}$ (and the requested values).
  • 1 $2x+3y=\sin y$
  • 2 $x^2+xy+y^2=100$
  • 3 $x^3+x^2y+xy^2+y^3=81$
  • 4 $y=\cos(x+y)$
  • 5 $\dfrac{x^2}{9}+\dfrac{y^2}{4}=1$; the slope at $\left(\sqrt5,\tfrac43\right)$
  • 6 Find the points of the circle $x^2+y^2-2x-3=0$ where the tangent is horizontal, and where it is vertical.
  • 7 $\sqrt x+\sqrt y=\sqrt{10}$; the slope at $(2.5,\,2.5)$
1 $2+3y'=\cos y\,y'$, so $y'=\dfrac{2}{\cos y-3}$ (the denominator is never $0$).
2 $2x+y+xy'+2yy'=0$, so $y'=-\dfrac{2x+y}{x+2y}$.
3 $3x^2+2xy+x^2y'+y^2+2xyy'+3y^2y'=0$, so $y'=-\dfrac{3x^2+2xy+y^2}{x^2+2xy+3y^2}$.
4 $y'=-\sin(x+y)(1+y')$, so $y'\big(1+\sin(x+y)\big)=-\sin(x+y)$ and $y'=-\dfrac{\sin(x+y)}{1+\sin(x+y)}$.
5 $\dfrac{2x}{9}+\dfrac{2yy'}{4}=0$ gives $y'=-\dfrac{4x}{9y}$. The point is on the curve ($\tfrac59+\tfrac49=1$). Slope $=-\dfrac{4\sqrt5}{9\cdot\frac43}=-\dfrac{\sqrt5}{3}\approx-0.745$.
6 $2x+2yy'-2=0$, so $y'=\dfrac{1-x}{y}$. Horizontal: $x=1$, then $y^2=4$: $(1,2)$ and $(1,-2)$. Vertical: $y=0$, then $x^2-2x-3=0$: $(3,0)$ and $(-1,0)$. (The circle has centre $(1,0)$ and radius $2$ — the answers are its top, bottom, right and left points.)
7 $\dfrac{1}{2\sqrt x}+\dfrac{y'}{2\sqrt y}=0$, so $y'=-\sqrt{\dfrac yx}$. At $(2.5,2.5)$ (on the curve, since $2\sqrt{2.5}=\sqrt{10}$): $y'=-1$.
IV
Inverse trigonometric functions. Differentiate (on the stated interval), or prove.
  • 1 $\cos^{-1}(x^2)$, $0\lt x\lt1$
  • 2 $\tan^{-1}(3x)$
  • 3 $\sin^{-1}\sqrt x$, $0\lt x\lt1$
  • 4 $x\sin^{-1}x$, $-1\lt x\lt1$
  • 5 $\cos^{-1}\!\left(\dfrac{1-x^2}{1+x^2}\right)$, $0\lt x\lt1$
  • 6 $\tan^{-1}\sqrt{\dfrac{1-\cos x}{1+\cos x}}$, $0\lt x\lt\pi$
  • 7 Prove that $\dfrac{d}{dx}\cos^{-1}x=-\dfrac{1}{\sqrt{1-x^2}}$ for $-1\lt x\lt1$.
  • 8 Prove that $\dfrac{d}{dx}\sec^{-1}x=\dfrac{1}{x\sqrt{x^2-1}}$ for $x\gt1$.
1 $-\dfrac{1}{\sqrt{1-x^4}}\cdot2x=-\dfrac{2x}{\sqrt{1-x^4}}$
2 $\dfrac{3}{1+9x^2}$
3 $\dfrac{1}{\sqrt{1-x}}\cdot\dfrac{1}{2\sqrt x}=\dfrac{1}{2\sqrt{x(1-x)}}$
4 $\sin^{-1}x+\dfrac{x}{\sqrt{1-x^2}}$
5 $x=\tan\theta$, $0\lt\theta\lt\tfrac\pi4$: $\dfrac{1-\tan^2\theta}{1+\tan^2\theta}=\cos2\theta$ with $0\lt2\theta\lt\tfrac\pi2$, inside $[0,\pi]$. So $y=2\theta=2\tan^{-1}x$ and $y'=\dfrac{2}{1+x^2}$.
6 $\dfrac{1-\cos x}{1+\cos x}=\dfrac{2\sin^2(x/2)}{2\cos^2(x/2)}=\tan^2\dfrac x2$, and $\tan\dfrac x2\gt0$ for $0\lt x\lt\pi$, so the square root is $\tan\dfrac x2$. With $\dfrac x2\in\left(0,\dfrac\pi2\right)$, $y=\dfrac x2$ and $y'=\dfrac12$.
7 $y=\cos^{-1}x$: $\cos y=x$ with $0\lt y\lt\pi$. Then $-\sin y\,y'=1$. On $(0,\pi)$, $\sin y\gt0$, so $\sin y=\sqrt{1-x^2}$ and $y'=-\dfrac{1}{\sqrt{1-x^2}}$.
8 $y=\sec^{-1}x$, $x\gt1$: $\sec y=x$ with $0\lt y\lt\tfrac\pi2$. Then $\sec y\tan y\,y'=1$; here $\tan y\gt0$ and $\tan y=\sqrt{\sec^2y-1}=\sqrt{x^2-1}$. So $y'=\dfrac{1}{x\sqrt{x^2-1}}$.
🎯 Exam corner Board · 1 markBoard · 2 marksBoard · 3 marksJEE MainJEE Advanced

How this is examined: Board papers ask for $\tfrac{dy}{dx}$ of composite, implicit and inverse-trigonometric functions (often with a substitution such as $x=\tan\theta$ to simplify first), and for a proof that a given function is or is not differentiable at a point. JEE adds modulus-type functions and numerical-value questions.

E1. Board · 1 mark $\dfrac{d}{dx}\sin^{-1}(x^2)$, for $0\lt x\lt1$, equals  (A) $\dfrac{1}{\sqrt{1-x^4}}$  (B) $\dfrac{2x}{\sqrt{1-x^4}}$  (C) $\dfrac{2x}{\sqrt{1-x^2}}$  (D) $-\dfrac{2x}{\sqrt{1-x^4}}$

Idea Chain rule: $\dfrac{1}{\sqrt{1-(x^2)^2}}\cdot2x$.
Answer (B). Trap (A) forgets the inside derivative $2x$; (C) forgets to square the inside under the root.

E2. Board · 1 mark Assertion–Reason. Assertion (A): $f(x)=|x-3|$ is continuous at $x=3$ but not differentiable there. Reason (R): every function differentiable at a point is continuous at that point.  (A) Both A and R are true and R is the correct explanation of A  (B) Both A and R are true but R is not the correct explanation of A  (C) A is true but R is false  (D) A is false but R is true

Idea A is true ($Lf'(3)=-1$, $Rf'(3)=1$) and R is Theorem 5.4. But R goes from differentiable to continuous; it cannot explain why a continuous function fails to be differentiable.
Answer (B).

E3. Board · 2 marks If $x^2+xy+y^2=7$, find $\dfrac{dy}{dx}$ at the point $(1,2)$.

Differentiate $2x+\left(y+x\dfrac{dy}{dx}\right)+2y\dfrac{dy}{dx}=0$, so $\dfrac{dy}{dx}=-\dfrac{2x+y}{x+2y}$. (1 mark)
Answer $(1,2)$ is on the curve ($1+2+4=7$). $\dfrac{dy}{dx}=-\dfrac{2+2}{1+4}=-\dfrac45$. (1 mark)

E4. Board · 3 marks Find $a$ and $b$ so that $f(x)=\begin{cases}x^2+3x+a,&x\le1\\ bx+2,&x\gt1\end{cases}$ is differentiable at $x=1$.

Continuity Differentiable $\Rightarrow$ continuous: $1+3+a=b+2$, i.e. $a-b=-2$. (1 mark)
Derivatives $Lf'(1)=\lim\limits_{h\to0^-}\dfrac{(1+h)^2+3(1+h)+a-(4+a)}{h}=\lim(5+h)=5$. Using continuity, $Rf'(1)=\lim\limits_{h\to0^+}\dfrac{b(1+h)+2-(b+2)}{h}=b$. (1 mark)
Answer $b=5$ and $a=3$. Check: both pieces give $7$ at $x=1$, and both slopes are $5$. (1 mark)

E5. JEE Main The number of points at which $f(x)=|x^2-1|+|x-2|$ is not differentiable is  (A) $1$  (B) $2$  (C) $3$  (D) $4$

Method $|x^2-1|$ has corners at $x=\pm1$ (simple zeros: the derivative $2x$ of $x^2-1$ is $\pm2\ne0$ there); $|x-2|$ has a corner at $x=2$. At each of these points the other term is differentiable, so each corner survives in the sum.
Check at 1 For $x$ near $1$, $f(x)=|x^2-1|+2-x$: $Rf'(1)=2-1=1$, $Lf'(1)=-2-1=-3$. ✓
Answer (C) $3$, at $x=-1,1,2$. Trap (B) misses $x=-1$.

E6. JEE Advanced (Numerical answer.) Let $y=\tan^{-1}\!\left(\dfrac{\sqrt{1+x^2}-1}{x}\right)$ for $x\ne0$. Find the value of $100\,\dfrac{dy}{dx}$ at $x=1$.

Substitute $x=\tan\theta$ with $\theta\in\left(-\tfrac\pi2,\tfrac\pi2\right)$, $\theta\ne0$. Then $\sqrt{1+x^2}=\sec\theta$ (positive) and $$\dfrac{\sec\theta-1}{\tan\theta}=\dfrac{1-\cos\theta}{\sin\theta}=\dfrac{2\sin^2(\theta/2)}{2\sin(\theta/2)\cos(\theta/2)}=\tan\dfrac\theta2.$$
Simplify $\tfrac\theta2\in\left(-\tfrac\pi4,\tfrac\pi4\right)$ is in the principal branch, so $y=\dfrac\theta2=\dfrac12\tan^{-1}x$ and $\dfrac{dy}{dx}=\dfrac{1}{2(1+x^2)}$.
Answer At $x=1$: $\dfrac14$, so $100\,\dfrac{dy}{dx}=\mathbf{25}$. ⏱ Fast route: a numerical difference quotient at $x=1$ gives $0.25$ directly — useful to confirm the simplification.
🔬 Research–Teaching NexusMini project · about 45 minutes

Research question. How fast does a bicycle move for a given pedalling rate — and can the chain rule predict it before you ride?

Why it matters. Gear trains are chains of rates: in bicycles, car gearboxes, wind turbines and the drive of a Mars rover, engineers multiply the rate of each stage to design for a target speed or torque. The chain rule is that multiplication. Cyclists in races choose gears by exactly this calculation.

What to do.

  1. On a geared bicycle, count the teeth on the front chainring ($T_f$) and the rear sprocket ($T_r$) in use, and measure the wheel diameter $D$ (in metres).
  2. Let $p$ be the number of pedal turns, $w$ the number of wheel turns and $s$ the distance travelled. Explain why $\dfrac{dw}{dp}=\dfrac{T_f}{T_r}$ and $\dfrac{ds}{dw}=\pi D$.
  3. Use the chain rule to write the speed $\dfrac{ds}{dt}=\dfrac{ds}{dw}\cdot\dfrac{dw}{dp}\cdot\dfrac{dp}{dt}$ in terms of the cadence $\dfrac{dp}{dt}$ (pedal turns per second). Tabulate the predicted speed in km/h for each gear at a cadence of $1.25$ turns per second ($75$ per minute).
  4. Test one or two gears: ride at a steady cadence (count turns for $20$ s) and record the speed with a phone GPS app or by timing a measured $100$ m. Compare with your prediction.

Tools: A geared bicycle, measuring tape, stopwatch or phone GPS app, spreadsheet.

Hand in: The tooth counts, the chain-rule formula, the table of predicted speeds, your measured speeds with percentage errors, and two sentences on the sources of error.

Going further. Real wheels have a slightly smaller rolling radius under load. Use your measurements to estimate the effective diameter, and say how much it changes your predictions.

Instructor note: with $T_f=44$, $T_r=11$, $D=0.68$ m and cadence $1.25$ turns/s, $\tfrac{ds}{dt}=\pi(0.68)(4)(1.25)\approx10.7$ m/s $\approx38$ km/h; with $T_r=22$ it halves to about $19$ km/h. Measured values are typically $2$–$5\%$ lower because of tyre compression and cadence drift.

🧪 Self-check quiz

0/5 correct

Q1. At $x=0$, the function $|x|$ is

Q2. $\dfrac{d}{dx}\sin(x^3)=$

Q3. If $x^2+y^2=9$, then $\dfrac{dy}{dx}=$

Q4. $\dfrac{d}{dx}\tan^{-1}(2x)=$

Q5. Which statement is always true?

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