Year 2 Pure · Chapter 12

Differentiation Rules

Specification: Edexcel Pure 7.1, 7.2, 7.4 — Differentiation · AQA section G1, G2, G4

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In this chapter

1. sin and cos from first principles
2. Exponentials and logarithms
3. The chain rule
4. The product rule
5. The quotient rule
6. Using dx/dy
Practice

Key facts — this chapter

With $x$ in radians, $\dfrac{d}{dx}(\sin x) = \cos x$ and $\dfrac{d}{dx}(\cos x) = -\sin x$ .
$\dfrac{d}{dx}(e^{kx}) = ke^{kx}$ , $\dfrac{d}{dx}(a^x) = a^x\ln a$ and $\dfrac{d}{dx}(\ln x) = \dfrac{1}{x}$ .
Chain rule: if $y$ is a function of $u$ and $u$ is a function of $x$ , then $\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}$ .
Product rule: $\dfrac{d}{dx}(uv) = u\dfrac{dv}{dx} + v\dfrac{du}{dx}$ .
Quotient rule: $\dfrac{d}{dx}\left(\dfrac{u}{v}\right) = \dfrac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$ .
$\dfrac{d}{dx}(\tan x) = \sec^2 x$ , and $\dfrac{dx}{dy} = 1 \div \dfrac{dy}{dx}$ .

In Year 1 you could differentiate powers of $x$ . This chapter adds the other functions, sines, exponentials and logarithms, and three rules for combining them: one for a function of a function, one for a product and one for a quotient. Together they differentiate almost anything you can write down.

1. sin and cos from first principles

The proof uses two limits. Both come from the small-angle approximations of Chapter 8, which is why the angle must be in radians.

Key result

With $h$ in radians, as $h \to 0$ ,

\frac{\sin h}{h} \to 1 \quad\text{and}\quad \frac{\cos h - 1}{h} \to 0

For small $h$ , $\sin h \approx h$ , and $\cos h - 1 \approx -\frac{1}{2}h^2$ , so the second fraction is about $-\frac{1}{2}h$ .

Worked example 12.1 · The derivative of sin from first principles

Prove from first principles that the derivative of $\sin x$ is $\cos x$ .

Expand $\sin(x + h)$ with the addition formula and collect terms:
$\begin{aligned} &\sin(x + h) - \sin x \\ &= \sin x\cos h + \cos x\sin h - \sin x \\ &= \sin x(\cos h - 1) + \cos x\sin h \end{aligned}$
Divide by $h$ :
$\begin{aligned} &\frac{\sin(x + h) - \sin x}{h} \\ &= \sin x \cdot \frac{\cos h - 1}{h} + \cos x \cdot \frac{\sin h}{h} \end{aligned}$
As $h \to 0$ , the limits above turn this into $\sin x \times 0 + \cos x \times 1 = \cos x$ .

The same method with $\cos(x + h) = \cos x\cos h - \sin x\sin h$ shows that the derivative of $\cos x$ is $-\sin x$ .

Top: y = sin x from 0 to 2 pi with tangents drawn at x = 0, pi over 2 and pi, of gradient 1, 0 and minus 1. Bottom: y = cos x, passing through the matching points (0, 1), (pi over 2, 0) and (pi, minus 1) — The gradient of sin x (top) at each point is the height of cos x (bottom) there.

Common mistake

These derivatives are only true in radians. In degrees the gradient of $\sin x$ at $x = 0$ is $\frac{\pi}{180}$ , not $1$ , and every result in this chapter would need an extra factor.

2. Exponentials and logarithms

Key result

\begin{aligned} \frac{d}{dx}(e^{kx}) &= ke^{kx} \\ \frac{d}{dx}(a^{x}) &= a^{x}\ln a \\ \frac{d}{dx}(\ln x) &= \frac{1}{x} \end{aligned}

The second follows from the first, because $a = e^{\ln a}$ , so $a^x = e^{x\ln a}$ : that is $e^{kx}$ with $k = \ln a$ . Section 6 shows where the third comes from. Since $\ln(kx) = \ln k + \ln x$ , the derivative of $\ln(kx)$ is also $\frac{1}{x}$ .

Worked example 12.2 · Standard derivatives

Differentiate $y = 3e^{2x} - 5\ln x + 2^x$ , and find the gradient of the curve at $x = 1$ .

Differentiate term by term:
$\frac{dy}{dx} = 6e^{2x} - \frac{5}{x} + 2^x\ln 2$
At $x = 1$ :
$\begin{aligned} \frac{dy}{dx} &= 6e^2 - 5 + 2\ln 2 \\ &= 40.7 \text{ (3 s.f.)} \end{aligned}$

3. The chain rule

$y = (3x^2 - 1)^5$ is a function of a function: first work out $u = 3x^2 - 1$ , then $y = u^5$ . A small change in $x$ changes $u$ at the rate $\frac{du}{dx}$ , and each change in $u$ changes $y$ at the rate $\frac{dy}{du}$ , so the rates multiply.

Key result

If $y$ is a function of $u$ , and $u$ is a function of $x$ , then

\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}

In words: differentiate the outside, keeping the inside as it is, then multiply by the derivative of the inside.

A chain of two machines: x goes into the first, which outputs u = g(x) at rate du/dx; u goes into the second, which outputs y = f(u) at rate dy/du. Below, the overall rate dy/dx is the product of the two rates — Two machines in a chain: the overall rate is the product of the two rates.

Worked example 12.3 · A power of a bracket

Differentiate $y = (3x^2 - 1)^5$ .

Let $u = 3x^2 - 1$ , so $y = u^5$ . Then $\frac{dy}{du} = 5u^4$ and $\frac{du}{dx} = 6x$ .
Multiply, and write the answer in terms of $x$ :
$\begin{aligned} \frac{dy}{dx} &= 5(3x^2 - 1)^4 \times 6x \\ &= 30x(3x^2 - 1)^4 \end{aligned}$

Worked example 12.4 · Chains with the new functions

Differentiate (a) $e^{x^2}$ , (b) $\ln(\cos x)$ and (c) $\sin^3 x$ .

Each time, differentiate the outside function, then multiply by the derivative of the inside.

(a) The outside is $e^u$ and the inside is $x^2$ : $e^{x^2} \times 2x = 2xe^{x^2}$ .

(b) The outside is $\ln u$ and the inside is $\cos x$ : $\dfrac{1}{\cos x} \times (-\sin x) = -\tan x$ .

(c) $\sin^3 x$ means $(\sin x)^3$ , so the outside is the cube: $3\sin^2 x \times \cos x = 3\sin^2 x\cos x$ .

A special case comes up so often that it’s worth knowing on its own: if the inside is $kx$ , the chain rule just multiplies by $k$ . So $\frac{d}{dx}(\sin kx) = k\cos kx$ and $\frac{d}{dx}(\cos kx) = -k\sin kx$ .

4. The product rule

Key result

If $u$ and $v$ are functions of $x$ ,

\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}

Think of $uv$ as the area of a rectangle with sides $u$ and $v$ . When $x$ grows a little, $u$ grows by $\delta u$ and $v$ by $\delta v$ , and the area gains three pieces: $u\,\delta v$ along one side, $v\,\delta u$ along the other, and a tiny corner $\delta u\,\delta v$ . Divide by $\delta x$ and let $\delta x \to 0$ : the corner vanishes, and what’s left is the product rule.

The rectangle has sides u = 1 + x and v = 1 + ½x², so its area is uv. Increase x by δx and the area gains two strips and a corner. Make δx smaller and watch the corner's share shrink to nothing.

Worked example 12.5 · Stationary points of a product

Find the $x$ -coordinates of the stationary points of $y = x^2e^{3x}$ .

With $u = x^2$ and $v = e^{3x}$ , $\frac{du}{dx} = 2x$ and $\frac{dv}{dx} = 3e^{3x}$ :
$\begin{aligned} \frac{dy}{dx} &= x^2 \times 3e^{3x} + e^{3x} \times 2x \\ &= xe^{3x}(3x + 2) \end{aligned}$
$e^{3x}$ is never $0$ , so $\frac{dy}{dx} = 0$ when $x = 0$ or $x = -\tfrac{2}{3}$ .

5. The quotient rule

Key result

If $u$ and $v$ are functions of $x$ ,

\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\dfrac{du}{dx} - u\dfrac{dv}{dx}}{v^2}

The order on top matters, because it’s a subtraction: start with $v$ .

Worked example 12.6 · The derivative of tan

Show that $\dfrac{d}{dx}(\tan x) = \sec^2 x$ .

Write $\tan x = \frac{\sin x}{\cos x}$ , with $u = \sin x$ and $v = \cos x$ :

\begin{aligned} &\frac{d}{dx}(\tan x) \\ &= \frac{\cos x\cos x - \sin x(-\sin x)}{\cos^2 x} \\ &= \frac{\cos^2 x + \sin^2 x}{\cos^2 x} \\ &= \frac{1}{\cos^2 x} = \sec^2 x \end{aligned}

The reciprocal functions follow in the same way, from the quotient rule or the chain rule:

The other trig derivatives

\begin{aligned} \frac{d}{dx}(\tan x) &= \sec^2 x \\ \frac{d}{dx}(\sec x) &= \sec x\tan x \\ \frac{d}{dx}(\operatorname{cosec} x) &= -\operatorname{cosec} x\cot x \\ \frac{d}{dx}(\cot x) &= -\operatorname{cosec}^2 x \end{aligned}

Worked example 12.7 · A maximum

Find the coordinates of the stationary point of $y = \dfrac{\ln x}{x}$ , $x > 0$ , and show that it is a maximum.

With $u = \ln x$ and $v = x$ :
$\begin{aligned} \frac{dy}{dx} &= \frac{x \times \frac{1}{x} - \ln x \times 1}{x^2} \\ &= \frac{1 - \ln x}{x^2} \end{aligned}$
This is $0$ when $\ln x = 1$ , so $x = e$ and $y = \frac{1}{e}$ .
$x^2$ is positive, so $\frac{dy}{dx}$ has the sign of $1 - \ln x$ : positive for $x < e$ and negative for $x > e$ . The curve rises then falls, so $\left(e, \tfrac{1}{e}\right)$ is a maximum.

6. Using dx/dy

Key result

\frac{dx}{dy} = \frac{1}{\;\dfrac{dy}{dx}\;}

So when $x$ is given as a function of $y$ , find $\frac{dx}{dy}$ and turn it upside down.

Worked example 12.8 · Turning the gradient upside down

A curve has equation $x = y^3 + y$ . Find the gradient of the curve at the point $(2, 1)$ .

$\frac{dx}{dy} = 3y^2 + 1$ , which is $4$ when $y = 1$ .
So $\frac{dy}{dx} = \frac{1}{4}$ at $(2, 1)$ .

Where the derivative of ln comes from

If $y = \ln x$ , then $x = e^y$ , so $\frac{dx}{dy} = e^y = x$ . Turning it upside down gives $\frac{dy}{dx} = \frac{1}{x}$ .

Practice

Twelve questions in three tiers. Try each one properly before opening the solution — the solutions show full method, not just answers.

Warm-up

1 [4 marks]

Differentiate each of these with respect to $x$ .

$e^{5x}$ [1]
$4\ln x$ [1]
$3^x$ [1]
$\sin 4x$ [1]

Show solution

(a) $5e^{5x}$ (b) $\dfrac{4}{x}$ (c) $3^x\ln 3$ (d) $4\cos 4x$

2 [2 marks]

Differentiate $y = (2x + 1)^6$ .

Show solution

By the chain rule, with inside $2x + 1$ :

\frac{dy}{dx} = 6(2x + 1)^5 \times 2 = 12(2x + 1)^5

3 [2 marks]

Differentiate $y = x^3\ln x$ .

Show solution

By the product rule, with $u = x^3$ and $v = \ln x$ :

\begin{aligned} \frac{dy}{dx} &= x^3 \times \frac{1}{x} + \ln x \times 3x^2 \\ &= x^2(1 + 3\ln x) \end{aligned}

4 [2 marks]

Differentiate $y = \dfrac{x}{x + 1}$ .

Show solution

By the quotient rule, with $u = x$ and $v = x + 1$ :

\begin{aligned} \frac{dy}{dx} &= \frac{(x + 1) \times 1 - x \times 1}{(x + 1)^2} \\ &= \frac{1}{(x + 1)^2} \end{aligned}

Standard

5 [3 marks]

Find the gradient of $y = e^{\sin x}$ at the point where $x = \pi$ .

Show solution

By the chain rule, $\dfrac{dy}{dx} = \cos x\,e^{\sin x}$ . At $x = \pi$ this is $\cos\pi \times e^0 = -1$ .

6 [6 marks]

Find the coordinates of the stationary points of $y = x^2e^{-x}$ , and determine their nature.

Show solution

By the product rule,

\begin{aligned} \frac{dy}{dx} &= x^2 \times (-e^{-x}) + e^{-x} \times 2x \\ &= xe^{-x}(2 - x) \end{aligned}

which is $0$ when $x = 0$ or $x = 2$ : the points $(0, 0)$ and $(2, 4e^{-2})$ .

Differentiating again, $\dfrac{d^2y}{dx^2} = e^{-x}(x^2 - 4x + 2)$ . At $x = 0$ this is $2 > 0$ , so $(0, 0)$ is a minimum. At $x = 2$ it is $-2e^{-2} < 0$ , so $(2, 4e^{-2})$ is a maximum.

7 [3 marks]

Show that $\dfrac{d}{dx}(\sec x) = \sec x\tan x$ .

Show solution

Write $\sec x = (\cos x)^{-1}$ and use the chain rule:

\begin{aligned} &\frac{d}{dx}(\cos x)^{-1} \\ &= -(\cos x)^{-2} \times (-\sin x) \\ &= \frac{1}{\cos x} \times \frac{\sin x}{\cos x} = \sec x\tan x \end{aligned}

8 [3 marks]

A curve has equation $x = y^2 - 4y$ . Find the gradient of the curve at the point $(-3, 1)$ .

Show solution

$\dfrac{dx}{dy} = 2y - 4$ , which is $-2$ when $y = 1$ . So $\dfrac{dy}{dx} = -\tfrac{1}{2}$ at $(-3, 1)$ .

9 [4 marks]

Find the equation of the tangent to $y = \ln(2x - 3)$ at the point where $x = 2$ .

Show solution

When $x = 2$ , $y = \ln 1 = 0$ . By the chain rule, $\dfrac{dy}{dx} = \dfrac{2}{2x - 3}$ , which is $2$ when $x = 2$ . The tangent is $y - 0 = 2(x - 2)$ , that is $y = 2x - 4$ .

Exam style

10 [6 marks]

f(x) = \frac{x^2 + 3}{x - 1}, \quad x > 1

Show that $f'(x) = \dfrac{(x - 3)(x + 1)}{(x - 1)^2}$ . [3]
Find the coordinates of the stationary point of $y = f(x)$ , and show that it is a minimum. [3]

Show solution

(a) By the quotient rule,

\begin{aligned} &f'(x) \\ &= \frac{(x - 1) \times 2x - (x^2 + 3) \times 1}{(x - 1)^2} \\ &= \frac{x^2 - 2x - 3}{(x - 1)^2} = \frac{(x - 3)(x + 1)}{(x - 1)^2} \end{aligned}

(b) For $x > 1$ , $f'(x) = 0$ only when $x = 3$ , and $f(3) = \tfrac{12}{2} = 6$ . The denominator and $x + 1$ are positive, so $f'(x)$ has the sign of $x - 3$ : negative just before $x = 3$ and positive just after. So $(3, 6)$ is a minimum.

11 [5 marks]

Given that $\dfrac{\sin h}{h} \to 1$ and $\dfrac{\cos h - 1}{h} \to 0$ as $h \to 0$ , prove from first principles that the derivative of $\cos x$ is $-\sin x$ .

Show solution

Using the addition formula,

\begin{aligned} &\cos(x + h) - \cos x \\ &= \cos x\cos h - \sin x\sin h - \cos x \\ &= \cos x(\cos h - 1) - \sin x\sin h \end{aligned}

Dividing by $h$ ,

\begin{aligned} &\frac{\cos(x + h) - \cos x}{h} \\ &= \cos x \cdot \frac{\cos h - 1}{h} - \sin x \cdot \frac{\sin h}{h} \end{aligned}

As $h \to 0$ this tends to $\cos x \times 0 - \sin x \times 1 = -\sin x$ .

12 [7 marks]

A hot drink cools so that its temperature $t$ minutes after it is made is $T = 20 + 70e^{-0.05t}$ °C.

State its temperature when it is made. [1]
Find the rate at which it is cooling after 10 minutes. [3]
Find when it is cooling at $1$ °C per minute, and its temperature then. [3]

Show solution

(a) At $t = 0$ , $T = 20 + 70 = 90$ °C.

(b) $\dfrac{dT}{dt} = 70 \times (-0.05)e^{-0.05t} = -3.5e^{-0.05t}$ . At $t = 10$ this is $-3.5e^{-0.5} = -2.12$ (3 s.f.), so it is cooling at $2.12$ °C per minute.

(c) $3.5e^{-0.05t} = 1$ gives $e^{-0.05t} = \tfrac{1}{3.5}$ , so

t = \frac{\ln 3.5}{0.05} = 25.1 \text{ minutes (3 s.f.)}

Then $T = 20 + 70 \times \tfrac{1}{3.5} = 40$ °C.