Mini-Lesson

Differentiation

5 missions. Guess, learn, then check. Your place is saved.

Mission 1Speed needs two moments. So what is your speed at one instant?
Mission 2Fold a box from a 30 cm square card: what corner cut holds the most?
Mission 3How do you differentiate $e^{s i n x}$ e^(sin x), a function inside a function?
Mission 4"The rate of increase of crime is falling." Is crime falling?
Mission 5A balloon's radius grows 1 cm/s. Why does its volume grow ever faster?

The specification points and sources are listed under the lesson: see Sources.

Mission 1

Speed needs two moments. So what is your speed at one instant?

Make a guess: Speed needs two moments. So what is your speed at one instant?

Differentiation · first principles

Chords close in on the tangent

The gradient of a curve at a point is the gradient of the tangent there.
A chord joins $P (x, f (x))$ P(x, f(x)) to a nearby point $Q (x + h, f (x + h))$ Q(x+h, f(x+h)). Its gradient is

\frac{f ( x + h ) - f ( x )}{h}

As $h \to 0$ h → 0, $Q$ Q slides towards $P$ P and the chord turns into the tangent. The tangent's gradient is the limit of the chord gradients.
We write it $\frac{d y}{d x}$ dy/dx or $f^{'} (x)$ f'(x): the rate of change of $y$ y with respect to $x$ x.

Worked example Differentiation · first principles

Chord gradients on $y = x^{2}$ y = x²

Question

P (3, 9)

lies on

y = x^{2}

. Find the gradients of the chords from

P

to the points with

x = 3 + h

for

h = 1

,

0.1

and

0.01

, and hence estimate the gradient at

P

.

1
Use the chord gradient $\frac{( 3 + h ) ^{2} - 9}{h}$ ((3+h)² − 9)/h for each $h$ h.
2
$h = 1$ h = 1: $\frac{16 - 9}{1} = 7$ (16 − 9)/1 = 7
3
$h = 0.1$ h = 0.1: $\frac{9.61 - 9}{0.1} = 6.1$ (9.61 − 9)/0.1 = 6.1
4
$h = 0.01$ h = 0.01: $\frac{9.0601 - 9}{0.01} = 6.01$ (9.0601 − 9)/0.01 = 6.01
5
The gradients approach 6. In fact $\frac{( 3 + h ) ^{2} - 9}{h} = 6 + h$ ((3+h)² − 9)/h = 6 + h, which tends to 6 as $h \to 0$ h → 0.

Answer

the chord gradients are 7, 6.1 and 6.01; the gradient at

P

is 6.

⚠️ Watch out

putting

h = 0

straight into

\frac{( 3 + h ) ^{2} - 9}{h}

gives

\frac{0}{0}

. Simplify first, then let

h \to 0

.

Quick check

?On

y = x^{3}

, what is the gradient of the chord from

x = 2

to

x = 2.1

?

Differentiation · first principles

The derivative as a limit

Definition:

f^{'} (x) = h \to 0 lim \frac{f ( x + h ) - f ( x )}{h}

Method: expand $f (x + h)$ f(x+h), subtract $f (x)$ f(x), divide by $h$ h, then let $h \to 0$ h → 0.
For $f (x) = x^{2}$ f(x) = x²: $\frac{( x + h ) ^{2} - x ^{2}}{h} = \frac{2 x h + h ^{2}}{h} = 2 x + h$ ((x+h)² − x²)/h = (2xh + h²)/h = 2x + h, which tends to $2 x$ 2x. So $\frac{d}{d x} (x^{2}) = 2 x$ d/dx(x²) = 2x.
You may see $δ x$ δ x in place of $h$ h: $\frac{d y}{d x} = lim_{δ x \to 0} \frac{δy}{δ x}$ dy/dx = lim_(δ x → 0) (δ y)/(δ x).

Worked example Differentiation · first principles

Proving the derivative of $x^{3}$ x³

Question

Prove from first principles that the derivative of

x^{3}

is

3 x^{2}

.

1
Expand: $(x + h)^{3} = x^{3} + 3 x^{2} h + 3 x h^{2} + h^{3}$ (x+h)³ = x³ + 3x²h + 3xh² + h³
2
Subtract $x^{3}$ x³: $f (x + h) - f (x) = 3 x^{2} h + 3 x h^{2} + h^{3}$ f(x+h) − f(x) = 3x²h + 3xh² + h³
3
Divide by $h$ h (with $h \neq = 0$ h ≠ 0): $\frac{f ( x + h ) - f ( x )}{h} = 3 x^{2} + 3 x h + h^{2}$ (f(x+h) − f(x))/h = 3x² + 3xh + h²
4
Let $h \to 0$ h → 0: the terms $3 x h$ 3xh and $h^{2}$ h² tend to 0, so $f^{'} (x) = lim_{h \to 0} (3 x^{2} + 3 x h + h^{2}) = 3 x^{2}$ f'(x) = lim_(h → 0)(3x² + 3xh + h²) = 3x²

Answer

\frac{d}{d x} (x^{3}) = 3 x^{2}

, as required.

⚠️ Watch out

a proof must start from the limit definition and write "

lim_{h \to 0}

" until the last step. Quoting

n x^{n - 1}

earns nothing here.

Calculate

?For

f (x) = 2 x^{2} - 3 x

, show that

\frac{f ( x + h ) - f ( x )}{h} = 4 x + 2 h - 3

. Hence find the gradient of the curve

y = f (x)

at

x = 5

. Give an exact answer.

Differentiation · first principles

First principles for $sin x$ sin x (A-level only)

Use radians. Expand with the compound-angle formula: $sin (x + h) = sin x cos h + cos x sin h$ sin (x+h) = sin x cos h + cos x sin h.
Regroup:

\frac{sin ( x + h ) - sin x}{h} = sin x (\frac{cos h - 1}{h}) + cos x (\frac{sin h}{h})

Small angles: $sin h \approx h$ sin h ≈ h and $cos h \approx 1 - \frac{h ^{2}}{2}$ cos h ≈ 1 − h²/2, so $\frac{s i n h}{h} \to 1$ (sin h)/h → 1 and $\frac{c o s h - 1}{h} \approx - \frac{h}{2} \to 0$ (cos h − 1)/h ≈ −h/2 → 0.
So $\frac{d}{d x} (sin x) = cos x$ d/dx(sin x) = cos x. The same method gives the derivative of $cos x$ cos x.

Quick check

?From first principles,

\frac{c o s ( x + h ) - c o s x}{h} = cos x (\frac{c o s h - 1}{h}) - sin x (\frac{s i n h}{h})

. What is its limit as

h \to 0

(

x

in radians)?

Calculate

?A ball rolls down a ramp so that it has travelled

s = 0.6 t^{2}

metres after

t

seconds. Show that its average speed from

t = 3

to

t = 3 + h

is

3.6 + 0.6 h

m/s. Hence find its speed at the instant

t = 3

. Give an exact answer.

m/s

Mission check

?Speed needs two moments. So what is your speed at one instant?

Mission 2

Fold a box from a 30 cm square card: what corner cut holds the most?

Make a guess: Fold a box from a 30 cm square card: what corner cut holds the most?

Differentiation · stationary points

The power rule

For any rational $n$ n:

\frac{d}{d x} (a x^{n}) = an x^{n - 1}

Differentiate a sum or difference term by term. A constant term differentiates to 0.
First rewrite as powers of $x$ x: $x = x^{\frac{1}{2}}$ √x = x^(1/2), $\frac{1}{x ^{2}} = x^{- 2}$ 1/x² = x⁻², $\frac{3}{x} = 3 x^{- \frac{1}{2}}$ 3/(√x) = 3x^(−1/2). Expand brackets and split fractions over a single power.
Notation: $\frac{d y}{d x}$ dy/dx or $f^{'} (x)$ f'(x). Differentiating again gives the second derivative $\frac{d ^{2} y}{d x ^{2}}$ d²y/dx² or $f^{''} (x)$ f''(x).

Worked example Differentiation · stationary points

Rewriting before differentiating

Question

Given

y = \frac{x ^{2} + 6}{x}

,

x > 0

, find

\frac{d y}{d x}

and the gradient of the curve at

x = 4

.

1
Split the fraction and write as powers: $y = x^{\frac{3}{2}} + 6 x^{- \frac{1}{2}}$ y = x^(3/2) + 6x^(−1/2)
2
Multiply by the power, then reduce the power by 1: $\frac{d y}{d x} = \frac{3}{2} x^{\frac{1}{2}} - 3 x^{- \frac{3}{2}}$ dy/dx = 3/2x^(1/2) − 3x^(−3/2)
3
Substitute $x = 4$ x = 4: $\frac{3}{2} (2) - 3 \times \frac{1}{8} = 3 - \frac{3}{8}$ 3/2(2) − 3 × 1/8 = 3 − 3/8

Answer

\frac{d y}{d x} = \frac{3}{2} x^{\frac{1}{2}} - 3 x^{- \frac{3}{2}}

; the gradient at

x = 4

is

\frac{21}{8} = 2.625

.

⚠️ Watch out

you cannot differentiate the top and bottom separately. Rewrite as a sum of powers first.

Quick check

?Find

\frac{d y}{d x}

when

y = \frac{8}{x ^{3}}

.

Differentiation · stationary points

Tangents and normals

The gradient of the curve at $x = a$ x = a is $m = f^{'} (a)$ m = f'(a). The tangent there has gradient $m$ m.
The normal is perpendicular to the tangent, so its gradient is $- \frac{1}{m}$ −1/m.
Use $y - y_{1} = m (x - x_{1})$ y − y₁ = m(x − x₁) with the point $(a, f (a))$ (a, f(a)) on the curve.
Example: $y = x^{2}$ y = x² at $(3, 9)$ (3, 9) has gradient 6, so the tangent is $y - 9 = 6 (x - 3)$ y − 9 = 6(x − 3), that is $y = 6 x - 9$ y = 6x − 9.

Calculate

?The curve

C

has equation

y = x^{3} - 3 x^{2} + 2

. The normal to

C

at the point where

x = 3

meets the

x

-axis at

Q

. Find the

x

-coordinate of

Q

. Give an exact answer.

Differentiation · stationary points

Stationary points and their nature

$f$ f is increasing where $f^{'} (x) > 0$ f'(x) > 0 and decreasing where $f^{'} (x) < 0$ f'(x) < 0.
At a stationary point $\frac{d y}{d x} = 0$ dy/dx = 0. Solve this to find the $x$ x-coordinates.
Classify with the second derivative: $\frac{d ^{2} y}{d x ^{2}} < 0$ d²y/dx² < 0 gives a maximum, $\frac{d ^{2} y}{d x ^{2}} > 0$ d²y/dx² > 0 a minimum. If it is 0, check the sign of $\frac{d y}{d x}$ dy/dx either side.
Optimisation: write the quantity in terms of one variable, differentiate, set the derivative to 0, classify, then answer in context.

Worked example Differentiation · stationary points

Finding and classifying stationary points

Question

Find the stationary points of

y = x^{3} - 6 x^{2} + 9 x + 1

and determine their nature.

1
Differentiate: $\frac{d y}{d x} = 3 x^{2} - 12 x + 9 = 3 (x - 1) (x - 3)$ dy/dx = 3x² − 12x + 9 = 3(x − 1)(x − 3)
2
Set to zero: $x = 1$ x = 1 or $x = 3$ x = 3, giving $y = 5$ y = 5 and $y = 1$ y = 1
3
Differentiate again: $\frac{d ^{2} y}{d x ^{2}} = 6 x - 12$ d²y/dx² = 6x − 12
4
At $x = 1$ x = 1: $6 - 12 = - 6 < 0$ 6 − 12 = −6 < 0, a maximum. At $x = 3$ x = 3: $18 - 12 = 6 > 0$ 18 − 12 = 6 > 0, a minimum.

Answer

(1, 5)

is a maximum and

(3, 1)

is a minimum.

⚠️ Watch out

give both coordinates. Many students find

x = 1

and

x = 3

and stop, losing the

y

-values.

Sort it

?For

f (x) = x^{3} - 12 x

, decide whether

f

is increasing or decreasing at each value of

x

.

Tap a statement, then tap the group it belongs in. Then press Check.

Calculate

?A closed cylindrical can has volume 500 cm³ and radius

r

cm. Show that its total surface area is

A = 2 π r^{2} + \frac{1000}{r}

. Use calculus to find the value of

r

that makes

A

a minimum. Give your answer to 3 significant figures.

cm

Mission check

?Fold a box from a 30 cm square card: what corner cut holds the most?

Mission 3

How do you differentiate $e^{s i n x}$ e^(sin x), a function inside a function?

Make a guess: How do you differentiate $e^{s i n x}$ e^(sin x), a function inside a function?

Differentiation · chain, product and quotient rules

Standard derivatives (A-level only)

$y$ y	$\frac{d y}{d x}$ dy/dx
$e^{k x}$ e^kx	$k e^{k x}$ ke^kx
$a^{k x}$ a^kx	$k a^{k x} ln a$ ka^kxln a
$ln x$ ln x (and $ln k x$ ln kx)	$\frac{1}{x}$ 1/x
$sin k x$ sin kx	$k cos k x$ kcos kx
$cos k x$ cos kx	$- k sin k x$ −ksin kx
$tan k x$ tan kx	$k sec^{2} k x$ ksec² kx

Angles must be in radians. The trig results are false in degrees.
$ln k x = ln k + ln x$ ln kx = ln k + ln x, so the constant $k$ k disappears when you differentiate.
Example: $\frac{d}{d x} (2^{3 x}) = 3 \times 2^{3 x} ln 2$ d/dx(2^3x) = 3 × 2^3xln 2.

Differentiation · chain, product and quotient rules

The chain rule (A-level only)

For a function of a function, $y = f (u)$ y = f(u) where $u = g (x)$ u = g(x):

\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x}

In words: differentiate the outer function, leave the inside alone, then multiply by the derivative of the inside.
Rates multiply along the chain: if $u$ u changes 2 times as fast as $x$ x, and $y$ y 3 times as fast as $u$ u, then $y$ y changes 6 times as fast as $x$ x.
Inverse functions: $\frac{d y}{d x} = 1 / \frac{d x}{d y}$ dy/dx = 1 / dx/dy. If $x = y^{3} + y$ x = y³ + y, then $\frac{d x}{d y} = 3 y^{2} + 1$ dx/dy = 3y² + 1, so $\frac{d y}{d x} = \frac{1}{3 y ^{2} + 1}$ dy/dx = 1/(3y² + 1).

Worked example Differentiation · chain, product and quotient rules

Using the chain rule (A-level only)

Question

Differentiate

y = (2 x^{3} - 1)^{5}

.

1
Name the inside: $u = 2 x^{3} - 1$ u = 2x³ − 1, so $y = u^{5}$ y = u⁵
2
Differentiate each link: $\frac{d y}{d u} = 5 u^{4}$ dy/du = 5u⁴ and $\frac{d u}{d x} = 6 x^{2}$ du/dx = 6x²
3
Multiply and put $u$ u back: $\frac{d y}{d x} = 5 (2 x^{3} - 1)^{4} \times 6 x^{2}$ dy/dx = 5(2x³ − 1)⁴ × 6x²

Answer

\frac{d y}{d x} = 30 x^{2} (2 x^{3} - 1)^{4}

⚠️ Watch out

5 (2 x^{3} - 1)^{4}

alone forgets to multiply by the derivative of the inside. Expanding the bracket first works but wastes time.

Quick check

?Find

\frac{d y}{d x}

when

y = e^{3 x^{2}}

.

Differentiation · chain, product and quotient rules

Product and quotient rules (A-level only)

Product rule: if $y = uv$ y = uv, then $\frac{d y}{d x} = u \frac{d v}{d x} + v \frac{d u}{d x}$ dy/dx = udv/dx + vdu/dx.
Quotient rule: if $y = \frac{u}{v}$ y = u/v, then $\frac{d y}{d x} = \frac{v \frac{d u}{d x} - u \frac{d v}{d x}}{v ^{2}}$ dy/dx = (vdu/dx − udv/dx)/v². The order on top matters.
The quotient rule on $\frac{s i n x}{c o s x}$ (sin x)/(cos x) gives $\frac{c o s ^{2} x + s i n ^{2} x}{c o s ^{2} x} = sec^{2} x$ (cos² x + sin² x)/(cos² x) = sec² x.
The chain or quotient rule gives $\frac{d}{d x} sec x = sec x tan x$ d/dxsec x = sec xtan x, $\frac{d}{d x} cosec x = - cosec x cot x$ d/dxcosec x = −cosec xcot x, $\frac{d}{d x} cot x = - cosec^{2} x$ d/dxcot x = −cosec² x.

Worked example Differentiation · chain, product and quotient rules

Product rule and a stationary point (A-level only)

Question

Find the

x

-coordinates of the stationary points of

y = x^{2} e^{3 x}

.

1
Set $u = x^{2}$ u = x² and $v = e^{3 x}$ v = e^3x: $\frac{d u}{d x} = 2 x$ du/dx = 2x and $\frac{d v}{d x} = 3 e^{3 x}$ dv/dx = 3e^3x
2
Product rule: $\frac{d y}{d x} = 3 x^{2} e^{3 x} + 2 x e^{3 x}$ dy/dx = 3x²e^3x + 2xe^3x
3
Factorise: $\frac{d y}{d x} = x e^{3 x} (3 x + 2)$ dy/dx = xe^3x(3x + 2)
4
Set to zero: $e^{3 x} > 0$ e^3x > 0 for all $x$ x, so $x = 0$ x = 0 or $3 x + 2 = 0$ 3x + 2 = 0

Answer

x = 0

and

x = - \frac{2}{3}

⚠️ Watch out

never divide by

x

to "cancel" it: that loses the solution

x = 0

. Factorise instead.

Calculate

?Find the gradient of the curve

y = \frac{2 x - 1}{x + 3}

at the point where

x = 1

. Give your answer as a decimal.

Quick check

?Find

\frac{d y}{d x}

when

y = ln 3 x + sin 4 x

(

x > 0

, in radians).

Calculate

?The curve

y = (x - 2) e^{2 x}

has one stationary point. Find its

x

-coordinate. Give your answer as a decimal.

Mission check

?How do you differentiate

e^{s i n x}

, a function inside a function?

Mission 4

"The rate of increase of crime is falling." Is crime falling?

Make a guess: "The rate of increase of crime is falling." Is crime falling?

Differentiation · concavity and inflection

Sketching the gradient function

To sketch $y = f^{'} (x)$ y = f'(x) from $y = f (x)$ y = f(x), read off the gradient of $f$ f as you move left to right.
Where $f$ f has a stationary point, $f^{'}$ f' crosses or touches the $x$ x-axis.
Where $f$ f is increasing, $f^{'}$ f' is above the axis; where $f$ f is decreasing, $f^{'}$ f' is below it.
Where $f$ f is steepest, $f^{'}$ f' has a turning point.
On a displacement–time graph, the gradient function is the velocity.

Quick check

?A cubic curve

y = f (x)

has a maximum at

x = - 1

, a minimum at

x = 3

and no other stationary points. Which describes the graph of

y = f^{'} (x)

?

Differentiation · concavity and inflection

Convex and concave (A-level only)

$f^{''} (x)$ f''(x) is the rate of change of the gradient.
If $f^{''} (x) > 0$ f''(x) > 0 on an interval, the gradient is increasing: the curve is convex there (it bends upwards, like a bowl).
If $f^{''} (x) < 0$ f''(x) < 0 on an interval, the gradient is decreasing: the curve is concave there (it bends downwards, like a cap).
A curve can rise and be concave at the same time: it is going up, but less and less steeply.

Differentiation · concavity and inflection

Points of inflection (A-level only)

A point of inflection is where the curve changes between concave and convex: $f^{''} (x) = 0$ f''(x) = 0 and $f^{''} (x)$ f''(x) changes sign.
$f^{''} (x) = 0$ f''(x) = 0 alone is not enough: $y = x^{4}$ y = x⁴ has $f^{''} (0) = 0$ f''(0) = 0, but $f^{''} (x) = 12 x^{2} \geq 0$ f''(x) = 12x² ≥ 0 on both sides, so $(0, 0)$ (0, 0) is a minimum.
If $f^{'} (x) = 0$ f'(x) = 0 there as well, it is a stationary point of inflection (like $y = x^{3}$ y = x³ at the origin). If $f^{'} (x) \neq = 0$ f'(x) ≠ 0, it is a non-stationary point of inflection.

Worked example Differentiation · concavity and inflection

Finding a point of inflection (A-level only)

Question

For

y = x^{3} - 6 x^{2} + 9 x + 1

, find the point of inflection, show that it is one, and state whether it is stationary.

1
Differentiate twice: $\frac{d y}{d x} = 3 x^{2} - 12 x + 9$ dy/dx = 3x² − 12x + 9 and $\frac{d ^{2} y}{d x ^{2}} = 6 x - 12$ d²y/dx² = 6x − 12
2
Solve $\frac{d ^{2} y}{d x ^{2}} = 0$ d²y/dx² = 0: $x = 2$ x = 2, and $y = 8 - 24 + 18 + 1 = 3$ y = 8 − 24 + 18 + 1 = 3
3
Check the sign change: at $x = 1$ x = 1, $\frac{d ^{2} y}{d x ^{2}} = - 6 < 0$ d²y/dx² = −6 < 0 (concave); at $x = 3$ x = 3, it is $6 > 0$ 6 > 0 (convex)
4
Check the gradient: at $x = 2$ x = 2, $\frac{d y}{d x} = 12 - 24 + 9 = - 3 \neq = 0$ dy/dx = 12 − 24 + 9 = −3 ≠ 0

Answer

(2, 3)

is a non-stationary point of inflection; the curve is concave for

x < 2

and convex for

x > 2

.

⚠️ Watch out

solving

\frac{d ^{2} y}{d x ^{2}} = 0

and stopping earns only part of the marks. You must show the sign change.

Calculate

?The curve

y = x^{3} - 9 x^{2} + 24 x - 5

has one point of inflection. Find its

y

-coordinate. Give an exact answer.

Sort it

?Is the origin a point of inflection on each curve?

Tap a statement, then tap the group it belongs in. Then press Check.

Quick check

?On which interval is the curve

y = x^{4} - 6 x^{2}

concave?

Calculate

?The curve

y = x e^{x}

has one point of inflection. Find its

x

-coordinate. Give an exact answer.

Mission check

?"The rate of increase of crime is falling." Is crime falling?

Mission 5

A balloon's radius grows 1 cm/s. Why does its volume grow ever faster?

Make a guess: A balloon's radius grows 1 cm/s. Why does its volume grow ever faster?

Differentiation · rates, implicit and parametric

Connected rates of change (A-level only)

When two quantities are linked by a formula, their rates are linked by the chain rule:

\frac{d A}{d t} = \frac{d A}{d r} \times \frac{d r}{d t}

Ripple: $A = π r^{2}$ A = π r² and $\frac{d r}{d t} = 2$ dr/dt = 2 cm/s. At $r = 5$ r = 5: $\frac{d A}{d t} = 2 π r \times 2 = 20 π \approx 62.8$ dA/dt = 2π r × 2 = 20π ≈ 62.8 cm²/s.
To construct a differential equation, turn words into a rate: " $N$ N grows at a rate proportional to $N$ N" becomes $\frac{d N}{d t} = k N$ dN/dt = kN, with $k > 0$ k > 0. "Decreases" needs a minus sign.

Calculate

?Water is poured into an inverted cone at a constant 6 cm³ per second. When the water is

h

cm deep, the surface radius is

\frac{h}{2}

cm. Show that

V = \frac{π h ^{3}}{12}

. Find the rate at which the depth is rising when

h = 4

. Give your answer to 3 significant figures.

cm/s

Quick check

?In a simple model, the rate of decrease of the radius

r

of a mint is inversely proportional to the square of the radius. Which differential equation fits (

k

a positive constant,

t

time)?

Differentiation · rates, implicit and parametric

Implicit differentiation (A-level only)

In an equation like $x^{2} + x y + y^{2} = 7$ x² + xy + y² = 7, $y$ y is not the subject. Differentiate every term with respect to $x$ x.
A term in $y$ y needs the chain rule: $\frac{d}{d x} (y^{3}) = 3 y^{2} \frac{d y}{d x}$ d/dx(y³) = 3y²dy/dx.
A product needs the product rule: $\frac{d}{d x} (x y) = x \frac{d y}{d x} + y$ d/dx(xy) = xdy/dx + y.
Then collect the $\frac{d y}{d x}$ dy/dx terms on one side and factorise.

Worked example Differentiation · rates, implicit and parametric

Gradient of an implicit curve (A-level only)

Question

Find the gradient of the curve

x^{2} + x y + y^{2} = 7

at the point

(1, 2)

.

1
Check the point: $1 + 2 + 4 = 7$ 1 + 2 + 4 = 7, so $(1, 2)$ (1, 2) is on the curve.
2
Differentiate term by term: $2 x + (x \frac{d y}{d x} + y) + 2 y \frac{d y}{d x} = 0$ 2x + (xdy/dx + y) + 2ydy/dx = 0
3
Collect: $\frac{d y}{d x} (x + 2 y) = - (2 x + y)$ dy/dx(x + 2y) = −(2x + y), so $\frac{d y}{d x} = - \frac{2 x + y}{x + 2 y}$ dy/dx = −(2x + y)/(x + 2y)
4
Substitute $(1, 2)$ (1, 2): $\frac{d y}{d x} = - \frac{4}{5}$ dy/dx = −4/5

Answer

the gradient at

(1, 2)

is

- \frac{4}{5} = - 0.8

.

⚠️ Watch out

differentiating

x y

as just

\frac{d y}{d x}

(no product rule) gives

- \frac{2}{5}

, a common lost mark.

Quick check

?Find the gradient of the circle

x^{2} + y^{2} = 25

at the point

(3, 4)

.

Differentiation · rates, implicit and parametric

Parametric differentiation (A-level only)

If $x$ x and $y$ y are both given in terms of a parameter $t$ t, use the chain rule:

\frac{d y}{d x} = \frac{d y}{d t} \div \frac{d x}{d t}

The answer is in terms of $t$ t. For a tangent or normal at a given $t$ t, find the point $(x, y)$ (x, y) and the gradient from the same $t$ t.

Worked example Differentiation · rates, implicit and parametric

Tangent to a parametric curve (A-level only)

Question

A curve has parametric equations

x = t^{2} + 1

,

y = t^{3} - 3 t

. Find the equation of the tangent at the point where

t = 2

.

1
Find the point: $x = 5$ x = 5, $y = 8 - 6 = 2$ y = 8 − 6 = 2
2
Differentiate each: $\frac{d x}{d t} = 2 t = 4$ dx/dt = 2t = 4 and $\frac{d y}{d t} = 3 t^{2} - 3 = 9$ dy/dt = 3t² − 3 = 9 at $t = 2$ t = 2
3
Divide: $\frac{d y}{d x} = \frac{9}{4}$ dy/dx = 9/4
4
Tangent: $y - 2 = \frac{9}{4} (x - 5)$ y − 2 = 9/4(x − 5), so $4 y = 9 x - 37$ 4y = 9x − 37

Answer

y = \frac{9}{4} x - \frac{37}{4}

, or

9 x - 4 y - 37 = 0

⚠️ Watch out

dividing the other way,

\frac{d x}{d t} \div \frac{d y}{d t}

, gives

\frac{4}{9}

: that is

\frac{d x}{d y}

.

Calculate

?A curve is given by

x = 2 sin t

,

y = cos 2 t

. The normal to the curve at the point where

t = \frac{π}{6}

meets the

x

-axis at

P

. Find the

x

-coordinate of

P

. Give your answer as a decimal.

Mission check

?A balloon's radius grows 1 cm/s. Why does its volume grow ever faster?

Recap

Big ideas

The derivative $\frac{d y}{d x} = f^{'} (x)$ dy/dx = f'(x) is the gradient of the tangent and the rate of change of $y$ y with respect to $x$ x.
It is the limit of chord gradients: $f^{'} (x) = lim_{h \to 0} \frac{f ( x + h ) - f ( x )}{h}$ f'(x) = lim_(h → 0) (f(x+h) − f(x))/h. Expand, subtract, divide by $h$ h, then let $h \to 0$ h → 0.
A-level only: with $sin h \approx h$ sin h ≈ h and $cos h \approx 1 - \frac{h ^{2}}{2}$ cos h ≈ 1 − h²/2, first principles gives $\frac{d}{d x} sin x = cos x$ d/dxsin x = cos x and $\frac{d}{d x} cos x = - sin x$ d/dxcos x = −sin x.
$\frac{d}{d x} (a x^{n}) = an x^{n - 1}$ d/dx(axⁿ) = anxⁿ⁻¹ for rational $n$ n: rewrite roots and fractions as powers first.
Tangent gradient $m = f^{'} (a)$ m = f'(a); normal gradient $- \frac{1}{m}$ −1/m; then $y - y_{1} = m (x - x_{1})$ y − y₁ = m(x − x₁).
Stationary points solve $\frac{d y}{d x} = 0$ dy/dx = 0; $\frac{d ^{2} y}{d x ^{2}} < 0$ d²y/dx² < 0 means a maximum and $> 0$ > 0 a minimum.
Learn the standard derivatives of $e^{k x}$ e^kx, $ln x$ ln x, $sin k x$ sin kx, $cos k x$ cos kx, $tan k x$ tan kx, with $x$ x in radians.
Chain rule: $\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x}$ dy/dx = dy/du × du/dx. Outer, keep the inside, times the inner derivative.
Product: $u \frac{d v}{d x} + v \frac{d u}{d x}$ udv/dx + vdu/dx. Quotient: $\frac{v \frac{d u}{d x} - u \frac{d v}{d x}}{v ^{2}}$ (vdu/dx − udv/dx)/v².
Stationary points of $f$ f are roots of $f^{'}$ f'; where $f$ f increases, $f^{'}$ f' is above the axis.
A-level only: $f^{''} (x) > 0$ f''(x) > 0 convex, $f^{''} (x) < 0$ f''(x) < 0 concave.
A-level only: a point of inflection needs $f^{''} (x) = 0$ f''(x) = 0 and a sign change; it is stationary only if $f^{'} (x) = 0$ f'(x) = 0 there too.
Connected rates: $\frac{d V}{d t} = \frac{d V}{d r} \times \frac{d r}{d t}$ dV/dt = dV/dr × dr/dt. Words like "proportional to" and "decreases" become a differential equation.
Implicit: differentiate every term with respect to $x$ x; $y^{n}$ yⁿ gives $n y^{n - 1} \frac{d y}{d x}$ nyⁿ⁻¹dy/dx and $x y$ xy needs the product rule.
Parametric: $\frac{d y}{d x} = \frac{d y}{d t} \div \frac{d x}{d t}$ dy/dx = dy/dt ÷ dx/dt.

🏆

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⭐⭐⭐

You've worked through Differentiation for AQA A-level Mathematics (7357). 🎉

Mission 1Speed needs two moments. So what is your speed at one instant?
Mission 2Fold a box from a 30 cm square card: what corner cut holds the most?
Mission 3How do you differentiate $e^{s i n x}$ e^(sin x), a function inside a function?
Mission 4"The rate of increase of crime is falling." Is crime falling?
Mission 5A balloon's radius grows 1 cm/s. Why does its volume grow ever faster?

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Specification points and sources for this lesson are listed at the bottom of the page.

Differentiation

Speed needs two moments. So what is your speed at one instant?

Chords close in on the tangent

Chord gradients on y=x2y = x²

The derivative as a limit

Proving the derivative of x3x³

First principles for sinxsin x (A-level only)

Fold a box from a 30 cm square card: what corner cut holds the most?

The power rule

Rewriting before differentiating

Tangents and normals

Stationary points and their nature

Finding and classifying stationary points

How do you differentiate esinxe^(sin x), a function inside a function?

Standard derivatives (A-level only)

The chain rule (A-level only)

Using the chain rule (A-level only)

Product and quotient rules (A-level only)

Product rule and a stationary point (A-level only)

"The rate of increase of crime is falling." Is crime falling?

Sketching the gradient function

Convex and concave (A-level only)

Points of inflection (A-level only)

Finding a point of inflection (A-level only)

A balloon's radius grows 1 cm/s. Why does its volume grow ever faster?

Connected rates of change (A-level only)

Implicit differentiation (A-level only)

Gradient of an implicit curve (A-level only)

Parametric differentiation (A-level only)

Tangent to a parametric curve (A-level only)

Big ideas

Mini-lesson complete!

📣 Smashed it? Share your score

Chord gradients on $y = x^{2}$ y = x²

Proving the derivative of $x^{3}$ x³

First principles for $sin x$ sin x (A-level only)

How do you differentiate $e^{s i n x}$ e^(sin x), a function inside a function?