Chapter 12 · Limits and Derivatives

12.4 Algebra of Derivatives

Four rules replace first principles: the power rule $\tfrac{d}{dx}x^n=nx^{n-1}$, the sum and constant-multiple rules, the product rule $(uv)'=u'v+uv'$ and the quotient rule. Together they differentiate every polynomial and every trigonometric function in a line or two.

Finding $\tfrac{d}{dx}(x^2)$ from first principles took four lines. Finding $\tfrac{d}{dx}(x^7-4x^3+2\sin x)$ or $\tfrac{d}{dx}\left(\tfrac{x^2}{1+\cos x}\right)$ the same way would take a page, with many chances to slip. Mathematicians do not work like that. Instead, they prove a few rules once, from first principles, and then build derivatives of complicated functions out of the derivatives of simple ones — in the same way that the algebra of limits (Theorem 12.2) builds limits.

In this section we prove the power rule for $x^n$, the rules for sums, differences and constant multiples, and then the two less obvious rules, for products and quotients. With the derivatives of $\sin x$ and $\cos x$ from Section 12.3 they give the derivatives of all polynomials, all rational functions and all six trigonometric functions. We finish with tangents, velocities and marginal costs.

The Power Rule and Polynomials

📐 Theorem 12.9 — The power rule

For every positive integer $n$, $$\dfrac{d}{dx}\left(x^n\right)=nx^{n-1}.$$

Proof. Fix a point $a$. By the second form of Definition 12.3 and Theorem 12.4, $$f'(a)=\lim_{x\to a}\dfrac{x^n-a^n}{x-a}=na^{n-1}.$$ Since $a$ was any point, $f'(x)=nx^{n-1}$. $\blacksquare$

A second proof, by the binomial theorem (Section 7.1). $(x+h)^n=x^n+{}^nC_1x^{n-1}h$ $+\,{}^nC_2x^{n-2}h^2+\dots+h^n$, so $$\begin{aligned}\dfrac{(x+h)^n-x^n}{h}&=nx^{n-1}+{}^nC_2x^{n-2}h\\&\quad+\dots+h^{n-1}.\end{aligned}$$ Every term after the first contains $h$, so the limit as $h\to0$ is $nx^{n-1}$.

📐 Theorem 12.10 — Sum, difference and constant-multiple rules

Let $u$ and $v$ be differentiable at $x$ and let $c$ be a constant. Then $u\pm v$ and $cu$ are differentiable at $x$, and $$(u\pm v)'=u'\pm v',\qquad(cu)'=c\,u'.$$

Proof. $\dfrac{[u(x+h)+v(x+h)]-[u(x)+v(x)]}{h}$ $=\dfrac{u(x+h)-u(x)}{h}+\dfrac{v(x+h)-v(x)}{h}$. Both fractions have limits, $u'(x)$ and $v'(x)$, so by the sum rule for limits the left side tends to $u'(x)+v'(x)$. Differences are the same with minus signs, and $\dfrac{cu(x+h)-cu(x)}{h}$ $=c\cdot\dfrac{u(x+h)-u(x)}{h}\to cu'(x)$. $\blacksquare$

Together with $\tfrac{d}{dx}(c)=0$, these rules differentiate any polynomial term by term: $$\begin{aligned}&\dfrac{d}{dx}\left(a_nx^n+\dots+a_2x^2+a_1x+a_0\right)\\&\quad=na_nx^{n-1}+\dots+2a_2x+a_1.\end{aligned}$$

✏️ Example 1 — Differentiating polynomials

Differentiate (a) $4x^5-3x^3+7x-2$ (b) $(x+1)(x-3)$ (c) $\dfrac{x^3}{3}+5$ (d) $\dfrac{x^4+2x^2}{x}$, $x\ne0$.

💡 Show solution▶

(a) $4\cdot5x^4-3\cdot3x^2+7\cdot1-0=20x^4-9x^2+7$.

(b) Expand first: $x^2-2x-3$, so the derivative is $2x-2$.

(d) Simplify first: for $x\ne0$ the function is $x^3+2x$, with derivative $3x^2+2$.

✏️ Example 2 — Slopes, tangents and horizontal tangents

For the curve $y=x^3-3x$: (a) find the slope at $x=2$ and the equation of the tangent there; (b) find the points where the tangent is horizontal.

💡 Show solution▶

(a) $\tfrac{dy}{dx}=3x^2-3$, which is $9$ at $x=2$. The point is $(2,\,8-6)=(2,2)$, so the tangent is $y-2=9(x-2)$, i.e. $y=9x-16$.

(b) Horizontal means slope $0$: $3x^2-3=0$, $x=\pm1$. The points are $(1,-2)$ and $(-1,2)$ — the bottom of the "valley" and the top of the "hill" of the graph.

Check A numerical slope at $x=2$: $\tfrac{f(2.001)-f(1.999)}{0.002}=9.000001$ ✓.

Left: $y=f(x)$ with its tangent at the red point. Right: the derivative $y=f'(x)$, traced as the point moves. Drag the red point along the curve (or use the slider, or Play). The height of the blue point on the right is the slope of the red tangent on the left. Watch where $f'$ crosses zero: there the tangent on the left is horizontal.

$f(x)$$x$

$f$ with its tangent (drag the red point)

$f'$ — the slope of that tangent

$x$—

$f(x)$—

tangent slope (measured)—

$f'(x)$ from the rules—

$f$ is…—

The Product and Quotient Rules

It is tempting to guess that the derivative of a product is the product of the derivatives. It is not: $x\cdot x=x^2$ has derivative $2x$, but $(x)'(x)'=1\cdot1=1$. The correct rule has two terms.

📐 Theorem 12.11 — Product and quotient rules

Let $u$ and $v$ be differentiable at $x$. Then $uv$ is differentiable at $x$ and $$(uv)'=u'v+uv'.$$ If also $v(x)\ne0$, then $\tfrac uv$ is differentiable at $x$ and $$\left(\dfrac uv\right)'=\dfrac{u'v-uv'}{v^2}.$$

Proof of the product rule. Add and subtract $u(x+h)v(x)$ in the numerator: $$\begin{aligned}\dfrac{u(x+h)v(x+h)-u(x)v(x)}{h}&=u(x+h)\cdot\dfrac{v(x+h)-v(x)}{h}\\&\quad+v(x)\cdot\dfrac{u(x+h)-u(x)}{h}.\end{aligned}$$ As $h\to0$ the two difference quotients tend to $v'(x)$ and $u'(x)$, and $u(x+h)\to u(x)$ (a differentiable function has no jump: $u(x+h)-u(x)=h\cdot\tfrac{u(x+h)-u(x)}{h}$ $\to0\cdot u'(x)=0$). By the algebra of limits the whole expression tends to $u(x)v'(x)+v(x)u'(x)$.

Proof of the quotient rule. $$\begin{aligned}&\dfrac{u(x+h)}{v(x+h)}-\dfrac{u(x)}{v(x)}\\&\quad=\dfrac{u(x+h)v(x)-u(x)v(x+h)}{v(x+h)v(x)}.\end{aligned}$$ Add and subtract $u(x)v(x)$ in the numerator and divide by $h$: $$\begin{aligned}\dfrac{1}{v(x+h)v(x)}\Big[&v(x)\dfrac{u(x+h)-u(x)}{h}\\&-u(x)\dfrac{v(x+h)-v(x)}{h}\Big].\end{aligned}$$ As $h\to0$ this tends to $\dfrac{v(x)u'(x)-u(x)v'(x)}{v(x)^2}$, using $v(x+h)\to v(x)\ne0$. $\blacksquare$

A memory aid for the quotient rule: "bottom times derivative of top, minus top times derivative of bottom, all over bottom squared". The order matters, because of the minus sign.

✏️ Example 3 — The product rule

Differentiate (a) $x^2\sin x$ (b) $(2x+1)(x^3-x)$, and check by expanding (c) $\sin x\cos x$.

💡 Show solution▶

(a) $u=x^2$, $v=\sin x$: $(uv)'=2x\sin x+x^2\cos x$.

(b) $u=2x+1$, $v=x^3-x$: $2(x^3-x)+(2x+1)(3x^2-1)$ $=2x^3-2x+6x^3+3x^2-2x-1$ $=8x^3+3x^2-4x-1$.

(b) Check Expanding first, $(2x+1)(x^3-x)=2x^4+x^3-2x^2-x$, whose derivative is $8x^3+3x^2-4x-1$ ✓.

✏️ Example 4 — The quotient rule

Differentiate (a) $\dfrac{x+1}{x-1}$ (b) $\dfrac{x^2}{x^2+1}$ (c) $\dfrac{\sin x}{x}$.

💡 Show solution▶

(a) $\dfrac{1\cdot(x-1)-(x+1)\cdot1}{(x-1)^2}=\dfrac{-2}{(x-1)^2}$, for $x\ne1$.

(b) $\dfrac{2x(x^2+1)-x^2\cdot2x}{(x^2+1)^2}=\dfrac{2x}{(x^2+1)^2}$.

Check (a) at $x=3$: the formula gives $-\tfrac{2}{4}=-0.5$; numerically $\tfrac{f(3.001)-f(2.999)}{0.002}\approx-0.5000001$ ✓.

Picture $u(x)v(x)$ as the area of a rectangle with sides $u=1+x$ and $v=1+x^2$. When $x$ grows by $\Delta x$, the area grows by a strip $v\,\Delta u$, a strip $u\,\Delta v$ and a small corner $\Delta u\,\Delta v$. Press Play to shrink $\Delta x$: after dividing by $\Delta x$, the corner vanishes and the two strips give $u'v+uv'$.

$x$1$\Delta x$0.5

$\Delta(uv)/\Delta x$—

$v\,\Delta u/\Delta x$—

$u\,\Delta v/\Delta x$—

corner $/\Delta x$—

$u'v+uv'$—

Negative Powers and the Other Trigonometric Functions

The quotient rule immediately extends the power rule and finishes the list of trigonometric derivatives.

📐 Theorem 12.12 — The power rule for negative integers

For every positive integer $n$ and every $x\ne0$, $$\dfrac{d}{dx}\left(x^{-n}\right)=-nx^{-n-1}.$$ So $\tfrac{d}{dx}x^m=mx^{m-1}$ holds for every integer $m$ (for $m=0$ both sides are $0$, taking $x\ne0$).

Proof. $x^{-n}=\dfrac{1}{x^n}$. By the quotient rule with $u=1$, $v=x^n$: $\dfrac{0\cdot x^n-1\cdot nx^{n-1}}{x^{2n}}=-nx^{n-1-2n}=-nx^{-n-1}$. $\blacksquare$

📐 Theorem 12.13 — Derivatives of the six trigonometric functions

With $x$ in radians, and wherever the function is defined,

$f(x)$	$\sin x$	$\cos x$	$\tan x$	$\cot x$	$\sec x$	$\operatorname{cosec}x$
$f'(x)$	$\cos x$	$-\sin x$	$\sec^2x$	$-\operatorname{cosec}^2x$	$\sec x\tan x$	$-\operatorname{cosec}x\cot x$

Proof. The first two are Theorem 12.8. For $\tan x=\tfrac{\sin x}{\cos x}$: $\dfrac{\cos x\cos x-\sin x(-\sin x)}{\cos^2x}=\dfrac{1}{\cos^2x}=\sec^2x$. For $\sec x=\tfrac1{\cos x}$: $\dfrac{0-1\cdot(-\sin x)}{\cos^2x}=\dfrac{1}{\cos x}\cdot\dfrac{\sin x}{\cos x}=\sec x\tan x$. The results for $\cot x=\tfrac{\cos x}{\sin x}$ and $\operatorname{cosec}x=\tfrac1{\sin x}$ follow in the same way (Exercise III). $\blacksquare$

A pattern helps the memory: the three "co-" functions have derivatives that start with a minus sign and swap each function for its co-function.

✏️ Example 5 — Negative powers and roots

Differentiate (a) $\dfrac{1}{x^3}$ (b) $\dfrac{3}{x^2}-\dfrac{2}{x}+4\sqrt x$ (c) $\dfrac{x^2-1}{x}$, each for $x\gt0$.

💡 Show solution▶

(a) $x^{-3}\to-3x^{-4}=-\dfrac{3}{x^4}$.

(b) $3x^{-2}-2x^{-1}+4x^{1/2}$ has derivative $-6x^{-3}+2x^{-2}+4\cdot\dfrac{1}{2\sqrt x}$ $=-\dfrac{6}{x^3}+\dfrac{2}{x^2}+\dfrac{2}{\sqrt x}$ (using $(\sqrt x)'=\tfrac{1}{2\sqrt x}$ from Section 12.3).

Check (c) by the quotient rule: $\dfrac{2x\cdot x-(x^2-1)}{x^2}=\dfrac{x^2+1}{x^2}=1+\dfrac1{x^2}$ ✓.

✏️ Example 6 — Mixing the rules

Differentiate (a) $x\tan x$ (b) $\dfrac{1+\sin x}{\cos x}$ (c) $x^3\cos x-\dfrac{\sin x}{x^2}$, and find the slope of $y=x\tan x$ at $x=\tfrac\pi4$.

💡 Show solution▶

(a) Product rule: $\tan x+x\sec^2x$. At $x=\tfrac\pi4$: $1+\tfrac\pi4\cdot2=1+\tfrac\pi2\approx2.571$.

(b) Quotient rule: $\dfrac{\cos x\cdot\cos x-(1+\sin x)(-\sin x)}{\cos^2x}=\dfrac{1+\sin x}{\cos^2x}$. (Or write the function as $\sec x+\tan x$: the derivative $\sec x\tan x+\sec^2x$ is the same thing.)

(c) $3x^2\cos x-x^3\sin x$ minus $\dfrac{x^2\cos x-2x\sin x}{x^4}$, i.e. $3x^2\cos x-x^3\sin x-\dfrac{x\cos x-2\sin x}{x^3}$.

⚠️ Common mistakes with the rules

1. $(uv)'=u'v'$. Wrong — test it on $x\cdot x$. Use $u'v+uv'$.

2. $\left(\tfrac uv\right)'=\tfrac{u'}{v'}$, or the numerator in the wrong order. The quotient rule is $\tfrac{u'v-uv'}{v^2}$; swapping the terms changes the sign of the answer.

3. Forgetting that constants differentiate to $0$ — but a constant factor stays: $(5x^3)'=15x^2$, while $(x^3+5)'=3x^2$.

4. Using the power rule on $\sin^nx$ blindly. $nx^{n-1}$ is for a power of $x$. $\sin^2x=\sin x\cdot\sin x$ needs the product rule: $2\sin x\cos x$, not $2\sin x$.

5. Using the quotient rule when simplifying is quicker. $\tfrac{x^4+2x^2}{x}=x^3+2x$; dividing out first saves work and errors.

6. Mixing up signs of the "co-" functions. $(\cos x)'=-\sin x$, $(\cot x)'=-\operatorname{cosec}^2x$, $(\operatorname{cosec}x)'=-\operatorname{cosec}x\cot x$; the others are positive.

📘 In the CBSE syllabus — school assessment only (not in the year-end board exam)

The chain rule is taught and tested in school, but not in the year-end board exam; JEE Main lists it (and JEE Advanced uses it throughout differential calculus). In Class 12 it becomes one of the most-used rules of all.

The chain rule. Many functions are composites (Section 2.4): $\sin2x$ is "double, then take the sine", and $(x^2+1)^5$ is "square and add $1$, then take the fifth power". Write $y=f(u)$ with $u=g(x)$. If $g$ is differentiable at $x$ and $f$ is differentiable at $u=g(x)$, then $y=f(g(x))$ is differentiable at $x$ and $$\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx},$$ that is, $\big(f(g(x))\big)'=f'\big(g(x)\big)\cdot g'(x)$: the derivative of the outer function, evaluated at the inner one, times the derivative of the inner function. Why rates multiply: if $u$ changes $3$ times as fast as $x$ and $y$ changes $2$ times as fast as $u$, then $y$ changes $2\times3=6$ times as fast as $x$.

Worked example 1. Differentiate $y=(2x+1)^5$.
Solution. Inner $u=2x+1$, $\tfrac{du}{dx}=2$; outer $y=u^5$, $\tfrac{dy}{du}=5u^4$. So $\tfrac{dy}{dx}=5(2x+1)^4\cdot2=10(2x+1)^4$. Check at $x=0$: the formula gives $10$; numerically $\tfrac{(1.0002)^5-(0.9998)^5}{0.0002}\approx10.000$ ✓. (Expanding $(2x+1)^5$ by the binomial theorem and differentiating term by term gives the same answer, much more slowly.)

Worked example 2. Differentiate (a) $\sin2x$ (b) $\sin(x^2)$.
Solution. (a) $\cos2x\cdot2=2\cos2x$. Check by the product rule: $\sin2x=2\sin x\cos x$ has derivative $2(\cos^2x-\sin^2x)=2\cos2x$ ✓. (b) The inner function is $x^2$: $\cos(x^2)\cdot2x=2x\cos(x^2)$ — not $\cos(x^2)$, which forgets the inner derivative, and not $\cos2x$.

Exercises

Powers and polynomials.

1 $x^7$
2 $5x^4-2x^3+x-9$
3 $x^{10}+10x$
4 $(3x-2)^2$
5 $x^2(x+3)$
6 $\dfrac{x^4+2x}{x}$, $x\ne0$
7 $\dfrac{1}{x^5}$, $x\ne0$
8 For $f(x)=x^3-6x^2+9x+1$, find $f'(x)$ and the values of $x$ where $f'(x)=0$.
9 For $f(x)=1+x+\dfrac{x^2}{2}+\dfrac{x^3}{3}+\dots+\dfrac{x^{50}}{50}$, find $f'(1)$.

💡 Show solutions▶

1 $7x^6$.

2 $20x^3-6x^2+1$.

3 $10x^9+10$.

4 $9x^2-12x+4$, so $18x-12$.

5 $x^3+3x^2$, so $3x^2+6x$.

6 $x^3+2$, so $3x^2$.

7 $-5x^{-6}=-\dfrac{5}{x^6}$.

8 $f'(x)=3x^2-12x+9=3(x-1)(x-3)$, which is $0$ at $x=1$ and $x=3$.

9 $f'(x)=1+x+x^2+\dots+x^{49}$ (each $\tfrac{x^k}{k}$ gives $x^{k-1}$). At $x=1$ that is fifty $1$'s: $f'(1)=50$.

Products and quotients.

1 $x^3\sin x$
2 $(x^2+1)(x-5)$
3 $x\cos x$
4 $\sin^2x$ (as $\sin x\cdot\sin x$)
5 $\dfrac{2x-1}{x+3}$
6 $\dfrac{1}{x^2+1}$
7 $\dfrac{x}{\sin x}$
8 $\dfrac{1+\cos x}{\sin x}$
9 $\dfrac{x^2+3x}{x-1}$
10 $\dfrac{\sin x}{1+\cos x}$, and simplify your answer

Check: your Card II answer against a numerical slope

Part:$x$1.5

numerical slope—

answer formula—

agree?—

💡 Show solutions▶

1 $3x^2\sin x+x^3\cos x$.

2 $2x(x-5)+(x^2+1)=3x^2-10x+1$.

3 $\cos x-x\sin x$.

4 $\cos x\sin x+\sin x\cos x=2\sin x\cos x=\sin2x$.

5 $\dfrac{2(x+3)-(2x-1)}{(x+3)^2}=\dfrac{7}{(x+3)^2}$.

6 $\dfrac{0-1\cdot2x}{(x^2+1)^2}=-\dfrac{2x}{(x^2+1)^2}$.

7 $\dfrac{\sin x-x\cos x}{\sin^2x}$.

8 $\dfrac{-\sin x\cdot\sin x-(1+\cos x)\cos x}{\sin^2x}$ $=\dfrac{-(\sin^2x+\cos^2x)-\cos x}{\sin^2x}$ $=-\dfrac{1+\cos x}{\sin^2x}$.

9 $\dfrac{(2x+3)(x-1)-(x^2+3x)}{(x-1)^2}=\dfrac{x^2-2x-3}{(x-1)^2}$.

10 $\dfrac{\cos x(1+\cos x)+\sin^2x}{(1+\cos x)^2}$ $=\dfrac{1+\cos x}{(1+\cos x)^2}=\dfrac{1}{1+\cos x}$.

III

Trigonometric functions and mixed rules.

1 Use the quotient rule to prove $\dfrac{d}{dx}(\cot x)=-\operatorname{cosec}^2x$.
2 Prove $\dfrac{d}{dx}(\operatorname{cosec}x)=-\operatorname{cosec}x\cot x$.
3 $3\tan x-2\sec x$
4 $x^2\tan x$
5 Find the slope of $y=\sin x+\cos x$ at $x=\tfrac\pi4$, and say what the tangent looks like there.
6 $\dfrac{1-\cos x}{1+\cos x}$

💡 Show solutions▶

1 $\dfrac{-\sin x\cdot\sin x-\cos x\cos x}{\sin^2x}$ $=\dfrac{-1}{\sin^2x}=-\operatorname{cosec}^2x$.

2 $\dfrac{0-1\cdot\cos x}{\sin^2x}=-\dfrac1{\sin x}\cdot\dfrac{\cos x}{\sin x}$ $=-\operatorname{cosec}x\cot x$.

3 $3\sec^2x-2\sec x\tan x$.

4 $2x\tan x+x^2\sec^2x$.

5 $y'=\cos x-\sin x=0$ at $\tfrac\pi4$: the tangent is horizontal (this is the top of the curve, $y=\sqrt2$).

6 $\dfrac{\sin x(1+\cos x)-(1-\cos x)(-\sin x)}{(1+\cos x)^2}$ $=\dfrac{2\sin x}{(1+\cos x)^2}$.

Applications.

1 A particle moves along a line so that $s=t^3-6t^2+9t$ metres after $t$ seconds. Find its velocity, its velocity at $t=2$, and the times when it is momentarily at rest.
2 Find the equation of the tangent to $y=x^3-2x^2+1$ at $x=2$.
3 Find the points on $y=x^3-12x$ where the tangent is horizontal.
4 The curve $y=ax^2+bx$ passes through $(1,5)$ and has slope $8$ there. Find $a$ and $b$.
5 A shop's revenue from selling $x$ units is $R(x)=x(200-0.5x)$ rupees. Find the marginal revenue $R'(x)$ and its value at $x=120$.

💡 Show solutions▶

1 $v=3t^2-12t+9$; $v(2)=12-24+9=-3\ \text{m/s}$ (moving back towards the start). $v=3(t-1)(t-3)=0$ at $t=1\ \text{s}$ and $t=3\ \text{s}$.

2 $y(2)=8-8+1=1$; $y'=3x^2-4x=4$ at $x=2$. Tangent: $y-1=4(x-2)$, i.e. $y=4x-7$.

3 $3x^2-12=0$, $x=\pm2$: the points $(2,-16)$ and $(-2,16)$.

4 $a+b=5$ and $y'(1)=2a+b=8$. Subtracting, $a=3$, so $b=2$.

5 $R=200x-0.5x^2$, $R'(x)=200-x$; $R'(120)=80$, i.e. about ₹80 extra revenue for the $121$st unit.

📘 School-assessment practice (the chain rule — see the 📘 box; not in the year-end board exam). Differentiate with respect to $x$.

1 $(3x-2)^4$
2 $\cos5x$
3 $\sin^2x$, treating it as $u^2$ with $u=\sin x$ (compare Mistake 4)
4 $\dfrac{1}{(x^2+1)^2}$
5 $\tan(x^2)$
6 Find the slope of the tangent to $y=(2x-3)^5$ at $x=2$.

💡 Show solutions▶

1 $4(3x-2)^3\cdot3=12(3x-2)^3$.

2 $-\sin5x\cdot5=-5\sin5x$.

3 $2u\cdot u'=2\sin x\cos x=\sin2x$ — the same as the product rule gives.

4 Write it as $u^{-2}$ with $u=x^2+1$: $-2u^{-3}\cdot2x=-\dfrac{4x}{(x^2+1)^3}$.

5 $\sec^2(x^2)\cdot2x=2x\sec^2(x^2)$.

6 $y'=5(2x-3)^4\cdot2=10(2x-3)^4$; at $x=2$: $10\cdot1^4=10$.

🎯 Exam corner Board · 1 markBoard · 2 marksBoard · 3 marksBoard · Case studyJEE MainJEE Advanced

How this is examined: Board papers ask for derivatives by the product and quotient rules (2–3 marks), often of trigonometric quotients that simplify nicely, and for a velocity or a tangent from a polynomial model. JEE uses the rules inside tangent and slope conditions, and combines them with sums of series.

E1. Board · 1 mark If $y=x\sin x$, then $\dfrac{dy}{dx}$ at $x=\tfrac\pi2$ is (A) $1$ (B) $0$ (C) $\tfrac\pi2$ (D) $-1$

💡 Show solution▶

Idea $y'=\sin x+x\cos x$; at $\tfrac\pi2$: $1+\tfrac\pi2\cdot0=1$.

Trap (C) is the value of $y$ at $\tfrac\pi2$; (B) uses $(uv)'=u'v'=1\cdot\cos x$.

Answer (A).

E2. Board · 2 marks Differentiate $\dfrac{x^2+1}{x^2-1}$ with respect to $x$.

💡 Show solution▶

Rule $\dfrac{2x(x^2-1)-(x^2+1)2x}{(x^2-1)^2}$. (1 mark)

Answer $\dfrac{-4x}{(x^2-1)^2}$, for $x\ne\pm1$. (1 mark)

E3. Board · 3 marks Differentiate $\dfrac{\sin x+\cos x}{\sin x-\cos x}$.

💡 Show solution▶

Rule With $u=\sin x+\cos x$, $u'=\cos x-\sin x$, $v=\sin x-\cos x$, $v'=\cos x+\sin x$: the numerator is $(\cos x-\sin x)(\sin x-\cos x)$ $-(\sin x+\cos x)^2$. (1 mark)

Simplify $=-(\sin x-\cos x)^2-(\sin x+\cos x)^2$ $=-2(\sin^2x+\cos^2x)=-2$. (1 mark)

Answer $\dfrac{-2}{(\sin x-\cos x)^2}$. (1 mark)

E4. Board · Case study A ball is thrown straight up from a height of $1.5\ \text{m}$; its height after $t$ seconds is $y=1.5+14t-4.9t^2$ metres. (i) Find its velocity $v(t)$. (1 mark) (ii) When does it reach its highest point (where $v=0$)? (1 mark) (iii) Find the greatest height. (2 marks)

💡 Show solution▶

(i) $v=\tfrac{dy}{dt}=14-9.8t\ \text{m/s}$. (1 mark)

(ii) $14-9.8t=0$ gives $t=\tfrac{14}{9.8}=\tfrac{10}{7}\approx1.43\ \text{s}$. (1 mark)

(iii) $y=1.5+14\cdot\tfrac{10}{7}-4.9\cdot\tfrac{100}{49}$ $=1.5+20-10=11.5\ \text{m}$. (2 marks)

E5. JEE Main If $f(x)=1+2x+3x^2+\dots+20x^{19}$, then $f'(1)$ equals (A) $2660$ (B) $2870$ (C) $210$ (D) $3080$

💡 Show solution▶

Method The term $kx^{k-1}$ has derivative $k(k-1)x^{k-2}$, so $f'(1)=\displaystyle\sum_{k=1}^{20}k(k-1)=\sum k^2-\sum k$.

Sums $\sum_{k=1}^{20}k^2=\tfrac{20\cdot21\cdot41}{6}=2870$ and $\sum_{k=1}^{20}k=210$, so $f'(1)=2660$.

Trap (B) is $\sum k^2$ alone; (C) is $f(1)$; (D) is $\sum k(k+1)$, from raising the power instead of lowering it.

Answer (A).

E6. JEE Advanced One or more options correct. The curve $y=x^3+ax^2+bx$ passes through $(1,-2)$ and has a horizontal tangent there. Then (A) $a=0$ (B) $b=-3$ (C) the tangent at $x=-1$ is also horizontal (D) the curve passes through $(-1,-2)$

💡 Show solution▶

Equations Through $(1,-2)$: $1+a+b=-2$. Horizontal tangent: $y'(1)=3+2a+b=0$. Subtracting, $2+a=2$, so $a=0$ and $b=-3$.

Options The curve is $y=x^3-3x$ with $y'=3x^2-3$, which is $0$ at $x=-1$ too, so (C) is true. $y(-1)=-1+3=2$, so the curve passes through $(-1,2)$, not $(-1,-2)$: (D) is false.

Answer (A), (B) and (C).

🔬 Research–Teaching NexusMini project · about 60 minutes

Research question. A packaging company folds open boxes from square cards of side $20\ \text{cm}$ by cutting a square of side $x$ from each corner. Where does the derivative of the volume become zero — and is that the box that holds the most?

Why it matters. Packaging, tin cans and shipping containers are designed to use as little material as possible for the capacity they need. Engineers find the best design by locating where the rate of change of volume (or cost) with respect to a dimension is zero. You can already compute that rate with the rules of this section; Class 12 will prove why the zero marks a maximum.

What to do.

Show that the volume is $V(x)=x(20-2x)^2\ \text{cm}^3$ for $0\lt x\lt10$.
Expand and differentiate: find $V'(x)$ and the value of $x$ in $(0,10)$ where $V'(x)=0$. Compute $V$ there.
Make boxes from card (or paper) with $x=1,2,3,10/3,4,5\ \text{cm}$. Measure each volume by filling it with rice or dried lentils and pouring into a measuring jug.
Tabulate $x$, the measured volume, the formula $V(x)$ and the sign of $V'(x)$. Plot $V$ against $x$.
Describe the shape of the graph near the point where $V'=0$.

Tools: Card, ruler, scissors, tape, rice or lentils, a measuring jug, a spreadsheet

Hand in: Your derivative working, the table and graph, a photo of your boxes, and a paragraph explaining what the sign of $V'(x)$ tells you on either side of the special value.

Going further. Repeat for a rectangular card of $20\ \text{cm}\times30\ \text{cm}$. Is the best cut still a sixth of the shorter side?

Instructor note: $V=4x^3-80x^2+400x$, $V'=12x^2-160x+400=4(3x-10)(x-10)$, so $V'=0$ at $x=\tfrac{10}{3}\approx3.33\ \text{cm}$ (and at $10$, outside the range), with $V=\tfrac{16000}{27}\approx592.6\ \text{cm}^3$. $V'\gt0$ for $x\lt\tfrac{10}3$ and $V'\lt0$ after. Measured volumes are usually $5$–$10\%$ low because of folds and tape. For the $20\times30$ card the best cut is $x=\tfrac{25-5\sqrt7}{3}\approx3.92\ \text{cm}$, not a sixth of $20$.

🧪 Self-check quiz

0/5 correct

Q1. $\dfrac{d}{dx}\left(x^6\right)$ is

Q2. The product rule says $(uv)'$ equals

Q3. $\dfrac{d}{dx}(\tan x)$ is

Q4. The derivative of $\dfrac{x}{x+1}$ at $x=0$ is

Q5. The slope of the tangent to $y=x^3$ at $x=-1$ is

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