Formulas · Mathematics

Limits and derivatives

7 formulas, each with worked examples. Revision sheet · Practise these formulas

Class 11

The limit of (xⁿ − aⁿ) ÷ (x − a)

x \to a lim \frac{x ^{n} - a ^{n}}{x - a} = n a^{n - 1}

What each letter means
$x$	the variable, moving closer and closer to a
$a$	the number that x approaches
$n$	the power

As x gets closer and closer to a, the fraction (xⁿ − aⁿ) ÷ (x − a) gets closer and closer to n aⁿ⁻¹.

Why it works

For a whole number n, xⁿ − aⁿ = (x − a)(xⁿ⁻¹ + xⁿ⁻²a + … + aⁿ⁻¹). Cancelling x − a leaves n terms, and as x → a each of them becomes aⁿ⁻¹, so the total becomes n aⁿ⁻¹.

When to use it

Limits of the form (xⁿ − aⁿ) ÷ (x − a), which give 0 ÷ 0 if x = a is put in directly, and the derivative of xⁿ.

How to use it

Check that the limit has the form (xⁿ − aⁿ) ÷ (x − a); write plain numbers as powers (32 = 2⁵).
Read n and a.
Write n aⁿ⁻¹.

To remember: Bring the power down and lower it by one.

Other forms

Special case $lim_{x \to 1} \frac{x ^{n} - 1}{x - 1} = n$ When a = 1.
To remember itThis limit is the slope of the curve y = xⁿ at x = a, which is why the derivative of xⁿ is nxⁿ⁻¹.

Worked examples

Story

A square metal plate of side 3 cm is heated and grows. As its side x gets close to 3, the change in area per cm of change in side, (x² − 9) ÷ (x − 3), gets close to what?

This is the form with n = 2 and a = 3.
$2 \times 3 = 6$

Answer: 6 cm² for each cm of side

Picture

The chord of y = x³ from (1, 1) to a nearby point (x, x³) has slope (x³ − 1) ÷ (x − 1). What does this slope approach as x → 1?

This is the form with n = 3 and a = 1.
$3 \times 1^{2} = 3$
That is the slope of the tangent to the curve at (1, 1).

Answer: 3

Direct

Evaluate the limit of (x⁵ − 32) ÷ (x − 2) as x → 2.

Here 32 = 2⁵, so this is the form with n = 5 and a = 2.
$5 \times 2^{4} = 80$

Answer: 80

Reverse

The limit of (xⁿ − 1) ÷ (x − 1) as x → 1 is 9. Find n.

Try it first, then show the working

The limit is n × 1ⁿ⁻¹, which is just n; therefore n = 9.
$9 \times 1^{8} = 9$

Answer: n = 9

Exam

Evaluate the limit of (x¹⁵ − 1) ÷ (x¹⁰ − 1) as x → 1.

Try it first, then show the working

Divide the top and the bottom by x − 1 and use the rule on each: the top tends to 15 and the bottom to 10.
$\frac{15 \times 1 ^{14}}{10 \times 1 ^{9}} = \frac{3}{2}$

Answer: 3/2

Common mistake: Putting x = a straight in, which gives 0 ÷ 0, or writing aⁿ instead of n aⁿ⁻¹.

Class 11

The limit of sin x ÷ x

x \to 0 lim \frac{sin x}{x} = 1

What each letter means
$x$	an angle in radians, moving closer and closer to 0

For a small angle x in radians, sin x is almost exactly x, so sin x ÷ x gets closer and closer to 1 as x → 0.

Why it works

In a circle of radius 1, an angle x (in radians) cuts an arc x long. The half-chord sin x is a little shorter than the arc and tan x a little longer, so cos x < sin x ÷ x < 1. As x → 0, cos x → 1, and the fraction is squeezed to 1.

When to use it

Limits with sin or tan of a small angle, and the derivatives of sin x and cos x.

How to use it

Make the angle inside sin match the bottom: sin 4x ÷ 4x → 1.
Multiply and divide by the number you need to make them match.
Keep the angle in radians.

To remember: For small angles, sin x is nearly x.

Other forms

Special case $lim_{x \to 0} \frac{t a n x}{x} = 1$ When tan in place of sin, since tan x = sin x ÷ cos x and cos x → 1.
Special case $lim_{x \to 0} \frac{1 - c o s x}{x} = 0$ When the companion limit for cos, used for the derivative of cos x.

Worked examples

Story

A pendulum swings through a small angle of 0.05 radians, and engineers use sin x ≈ x for small swings. How good is this?

The ratio sin 0.05 ÷ 0.05 is about 0.9996, within 0.1% of 1, as the limit promises:
$lim_{x \to 0} \frac{s i n x}{x} = 1$

Answer: Very good: sin 0.05 ÷ 0.05 is about 0.9996

Picture

In a circle of radius 1, a small angle x at the centre cuts an arc x long and a chord 2 sin(x/2) long. What does chord ÷ arc approach as x → 0?

Chord ÷ arc = sin(x/2) ÷ (x/2), the same form with x/2 in place of x.
$lim_{x \to 0} \frac{2 s i n \frac{x}{2}}{x} = 1$

Answer: 1

Direct

Evaluate the limit of sin 4x ÷ sin 2x as x → 0.

Write it as (sin 4x ÷ 4x) × (2x ÷ sin 2x) × 2; the first two parts tend to 1.
$1 \times 1 \times 2 = 2$

Answer: 2

Reverse

The limit of sin kx ÷ x as x → 0 is 5. Find k.

Try it first, then show the working

Here sin kx ÷ x = k × (sin kx ÷ kx), which tends to k; therefore k = 5.
$lim_{x \to 0} \frac{s i n 5 x}{x} = 5$

Answer: k = 5

Exam

Evaluate the limit of (1 − cos 2x) ÷ x² as x → 0.

Try it first, then show the working

Use 1 − cos 2x = 2 sin² x: the fraction is 2 × (sin x ÷ x)².
$2 \times 1^{2} = 2$

Answer: 2

Common mistake: Working in degrees (then the limit is π/180, not 1), or cancelling the x in sin x ÷ x as if sin were a number.

Class 11

Derivative of xⁿ (the power rule)

\frac{d}{d x} x^{n} = n x^{n - 1}

What each letter means
$x$	the variable
$n$	the power (any rational number)

The derivative of xⁿ is nxⁿ⁻¹: bring the power down in front and lower the power by one.

Why it works

The derivative at x = a is the limit of (xⁿ − aⁿ) ÷ (x − a) as x → a, which is n aⁿ⁻¹ by the standard limit. Writing x for a gives nxⁿ⁻¹.

When to use it

Differentiating powers of x, polynomials (term by term), roots (√x = x^½) and reciprocals (1/x = x⁻¹); slopes of tangents and rates of change.

How to use it

Write each term as a power of x: √x = x^½, 1/x² = x⁻².
Multiply by the power, then take 1 off the power.
A constant on its own has derivative 0.

To remember: Power down, power minus one.

Other forms

Special case $\frac{d}{d x} (c x^{n}) = n c x^{n - 1}$ When the power is multiplied by a constant c.
Special case $\frac{d}{d x} x = 1$ When n = 1: the line y = x has slope 1.

Worked examples

Story

A square of side x cm has area x² cm². How fast does the area grow, per cm of side, when x = 5?

$\frac{d}{d x} x^{2} = 2 x$
$2 \times 5 = 10$

Answer: 10 cm² per cm

Picture

Find the slope of the tangent to y = x³ at x = 2.

$\frac{d}{d x} x^{3} = 3 x^{2}$
$3 \times 2^{2} = 12$

Answer: 12

Direct

Differentiate x⁷.

$\frac{d}{d x} x^{7} = 7 x^{6}$

Answer: 7x⁶

Reverse

The derivative of xⁿ is 6x⁵. Find n.

Try it first, then show the working

Matching nxⁿ⁻¹ with 6x⁵ gives n = 6.
$\frac{d}{d x} x^{6} = 6 x^{5}$

Answer: n = 6

Exam

Differentiate x⁻² + 3x⁴ − 5.

Try it first, then show the working

Take each term in turn; the constant 5 has derivative 0.
$\frac{d}{d x} (x^{- 2} + 3 x^{4} - 5) = - 2 x^{- 3} + 12 x^{3}$

Answer: −2x⁻³ + 12x³

Common mistake: Lowering the power without bringing it down in front, or raising the power by one (that is integration, in Class 12).

Class 11

The product rule

\frac{d}{d x} (uv) = u \frac{d v}{d x} + v \frac{d u}{d x}

What each letter means
$u$	a function of x
$v$	another function of x
$x$	the variable

The derivative of a product uv is the first times the derivative of the second, plus the second times the derivative of the first.

Why it works

When x changes by a little, u changes by Δu and v by Δv, and the product uv changes by uΔv + vΔu + ΔuΔv: the extra area of a rectangle with sides u and v. Dividing by the change in x and letting it shrink to 0, the last term vanishes, leaving u (dv/dx) + v (du/dx).

When to use it

Differentiating a product such as x² sin x or (x + 1)(x² − 3).

How to use it

Name the two factors u and v.
Find du/dx and dv/dx.
Write u × dv/dx + v × du/dx, and tidy.

To remember: First times the change in the second, plus second times the change in the first.

Other forms

Special case $\frac{d}{d x} (c u) = c \frac{d u}{d x}$ When v is a constant c (its derivative is 0).
To remember itFor a sum the rule is simpler: the derivative of u + v is du/dx + dv/dx.

Worked examples

Story

A rectangle has length 2x and width x + 3, both growing with x. How fast is its area growing when x = 1?

With the factors 2x and x + 3, the rate is 2x × 1 + (x + 3) × 2.
$2 \times 1 + (1 + 3) \times 2 = 10$

Answer: 10

Picture

Find the slope of the tangent to y = (x + 1)(x² − 3) at x = 2.

The slope is (x + 1) × 2x + (x² − 3) × 1; at x = 2 that is:
$(2 + 1) \times 4 + (4 - 3) = 13$

Answer: 13

Direct

Differentiate x² sin x.

Take the first factor x² and the second sin x.
$\frac{d}{d x} (x^{2} sin x) = x^{2} cos x + 2 x sin x$

Answer: x² cos x + 2x sin x

Reverse

The derivative of x times some function is x cos x + sin x. What is the function?

Try it first, then show the working

Compare with x (dv/dx) + v × 1: the function is sin x, whose derivative is cos x.
$\frac{d}{d x} (x sin x) = x cos x + sin x$

Answer: sin x

Exam

Differentiate sin x cos x.

Try it first, then show the working

Take the first factor sin x and the second cos x.
$\frac{d}{d x} (sin x cos x) = (cos x)^{2} - (sin x)^{2}$
That is cos 2x.

Answer: cos² x − sin² x, which is cos 2x

Common mistake: Multiplying the two derivatives: the derivative of uv is not (du/dx)(dv/dx).

Class 11

The quotient rule

\frac{d}{d x} (\frac{u}{v}) = \frac{v \frac{d u}{d x} - u \frac{d v}{d x}}{v ^{2}}

What each letter means
$u$	the function of x on the top
$v$	the function of x on the bottom (not 0)
$x$	the variable

The derivative of a fraction u/v is the bottom times the derivative of the top, minus the top times the derivative of the bottom, all over the bottom squared.

Why it works

Write u/v as u × (1/v) and use the product rule; the derivative of 1/v is −(dv/dx) ÷ v². Putting the two terms over v² gives the rule.

When to use it

Differentiating fractions such as (x + 1) ÷ (x − 1), tan x = sin x ÷ cos x, and 1 ÷ (x² + 1).

How to use it

Name the top u and the bottom v, and find their derivatives.
Work out v (du/dx) − u (dv/dx), in that order.
Divide by v².

To remember: Low d-high minus high d-low, over low squared.

Other forms

Special case $\frac{d}{d x} (\frac{1}{v}) = - \frac{\frac{d v}{d x}}{v ^{2}}$ When the top is 1.
To remember itLow d-high minus high d-low, all over the square of what is below.

Worked examples

Story

Making x items costs (100 + 2x) ÷ x rupees per item. How fast does the cost per item change when x = 10?

The top has derivative 2 and the bottom 1.
$\frac{10 \times 2 - ( 100 + 20 ) \times 1}{1 0 ^{2}} = - 1$

Answer: −1 rupee for each extra item

Picture

Find the slope of the tangent to y = 1 ÷ (x² + 1) at x = 1.

With the top 1, the slope is −2x ÷ (x² + 1)².
$\frac{- 2 \times 1}{( 1 + 1 ) ^{2}} = - \frac{1}{2}$

Answer: −1/2

Direct

Differentiate (x + 1) ÷ (x − 1).

The top and the bottom both have derivative 1.
$\frac{( x - 1 ) - ( x + 1 )}{( x - 1 ) ^{2}} = \frac{- 2}{( x - 1 ) ^{2}}$

Answer: −2 ÷ (x − 1)²

Reverse

The derivative of some function divided by x is (x cos x − sin x) ÷ x². What is the function?

Try it first, then show the working

Compare with (x × du/dx − u × 1) ÷ x²: the function is sin x, with derivative cos x.
$\frac{d}{d x} \frac{s i n x}{x} = \frac{x c o s x - s i n x}{x ^{2}}$

Answer: sin x

Exam

Use the quotient rule to show that the derivative of tan x is sec² x.

Try it first, then show the working

Write tan x = sin x ÷ cos x: the top has derivative cos x and the bottom −sin x.
$\frac{( c o s x ) ^{2} + ( s i n x ) ^{2}}{( c o s x ) ^{2}} = \frac{1}{( c o s x ) ^{2}}$
That is sec² x.

Answer: sec² x

Common mistake: Swapping the order on the top (u dv/dx − v du/dx gives the wrong sign), or forgetting to square the bottom.

Class 11

Derivative of sin x

\frac{d}{d x} sin x = cos x

What each letter means
$x$	an angle in radians

The slope of the sine curve at any point is the cosine of the angle there.

Why it works

From first principles, sin(x + h) − sin x = 2 cos(x + h/2) sin(h/2). Dividing by h gives cos(x + h/2) × (sin(h/2) ÷ (h/2)), and as h → 0 this tends to cos x × 1.

When to use it

Differentiating anything with sin x in it, and the rate of change of anything that swings or waves, like a pendulum or a tide.

How to use it

Keep x in radians.
Replace sin x by cos x; a number in front stays in front.

To remember: The derivatives go round in fours: sin, cos, −sin, −cos, and back to sin.

Other forms

To remember itThe sine curve is steepest (slope 1) where it crosses 0, and flat (slope 0) at its tops, exactly where cos x is 1 and 0.

Worked examples

Story

A float bobbing on water is sin t metres above its rest level at time t seconds. How fast is it rising at t = 0?

Its speed is the derivative, cos t.
$cos 0 = 1$

Answer: 1 m per second

Picture

Find the slope of the tangent to y = sin x at x = π/3.

$cos \frac{π}{3} = \frac{1}{2}$

Answer: 1/2

Direct

Differentiate 3 sin x + 2x.

$\frac{d}{d x} (3 sin x + 2 x) = 3 cos x + 2$

Answer: 3 cos x + 2

Reverse

Which function has derivative cos x and the value 0 at x = 0?

Try it first, then show the working

The derivative of sin x is cos x, and sin 0 = 0.
$sin 0 = 0$

Answer: sin x

Exam

Find the derivative of sin x from first principles.

Try it first, then show the working

The change is sin(x + h) − sin x = 2 cos(x + h/2) sin(h/2).
Dividing by h and letting h → 0 uses sin(h/2) ÷ (h/2) → 1, leaving cos x.
$lim_{h \to 0} \frac{s i n ( x + h ) - s i n x}{h} = cos x$

Answer: cos x

Common mistake: Working in degrees (the rule only holds for radians), or adding a minus sign (that belongs to cos x).

Class 11

Derivative of cos x

\frac{d}{d x} cos x = - sin x

What each letter means
$x$	an angle in radians

The slope of the cosine curve at any point is minus the sine of the angle there.

Why it works

From first principles, cos(x + h) − cos x = −2 sin(x + h/2) sin(h/2). Dividing by h and letting h → 0 gives −sin x × 1.

When to use it

Differentiating anything with cos x in it, and velocities of things that swing, like a spring or a pendulum.

How to use it

Keep x in radians.
Replace cos x by −sin x; a number in front stays in front.

To remember: Cos starts at its top and falls, so its slope starts negative: −sin.

Other forms

To remember itDerivatives that start with "co" (cos, cot, cosec) get a minus sign.

Worked examples

Story

A spring's end is at cos t cm from its rest point at time t seconds. Find its velocity at t = π/2.

The velocity is the derivative, −sin t.
$- sin \frac{π}{2} = - 1$

Answer: −1 cm per second (moving back)

Picture

Find the slope of the tangent to y = cos x at x = π/6.

$- sin \frac{π}{6} = - \frac{1}{2}$

Answer: −1/2

Direct

Differentiate 5 cos x − x².

$\frac{d}{d x} (5 cos x - x^{2}) = - 5 sin x - 2 x$

Answer: −5 sin x − 2x

Reverse

Which of sin x, cos x and −cos x has derivative sin x?

Try it first, then show the working

The derivative of cos x is −sin x, so the derivative of −cos x is sin x.
$\frac{d}{d x} (- cos x) = sin x$

Answer: −cos x

Exam

Differentiate sin x + cos x, and find where the slope is 0 for x between 0 and π/2.

Try it first, then show the working

$\frac{d}{d x} (sin x + cos x) = cos x - sin x$
The slope is 0 where cos x = sin x, that is at x = π/4.
$cos \frac{π}{4} - sin \frac{π}{4} = 0$

Answer: cos x − sin x; the slope is 0 at x = π/4

Common mistake: Forgetting the minus sign.