Differentiation in P3

A2 · P3 · 14 min

In P1 you could only differentiate powers of $x$ . P3 widens the toolkit to every function on the syllabus: $e^x$ , $\ln x$ , the six trigonometric functions and $\tan^{-1} x$ , combined in sums, products, quotients and composites, and extended to curves given implicitly or parametrically. This note is the map: it collects the standard derivatives, sharpens the chain rule (the one rule used in almost every P3 calculus question), and shows how gradients, tangents, normals and stationary points are examined now that the functions are harder. Differentiation appears on every P3 paper, often as a whole question and always as a step inside integration, numerical methods and differential equations questions.

What changes from P1

Nothing about the meaning of a derivative changes. $\dfrac{dy}{dx}$ is still the gradient of the curve, the rate at which $y$ changes as $x$ changes. What changes is the range of functions you can feed in and the number of rules you combine.

P1	P3
$x^n$ for rational $n$	$x^n$ , $e^x$ , $\ln x$ , $\sin x$ , $\cos x$ , $\tan x$ , $\sec x$ , $\operatorname{cosec} x$ , $\cot x$ , $\tan^{-1} x$
Chain rule for $(ax+b)^n$ and similar	Chain rule for any composite, including layered ones such as $\ln(\cos 2x)$
No products or quotients	Product rule and quotient rule
Curves $y = f(x)$ only	Implicit curves such as $x^2 + y^2 = xy + 7$ and parametric curves $x = f(t)$ , $y = g(t)$
Answers usually integers or fractions	Answers usually exact: $\ln 3$ , $e^{-2}$ , $\tfrac{\pi}{4}$ , $\sqrt{3}$

The dedicated notes cover each new piece in depth:

The standard derivatives

Every derivative in P3 is built from this table plus the chain, product and quotient rules. Learn the left-hand block by heart; the right-hand block follows from it, and you should be able to derive any line in under a minute.

Standard derivatives

$f(x)$	$f'(x)$	$f(x)$	$f'(x)$
$x^n$	$nx^{n-1}$	$\tan x$	$\sec^2 x$
$e^x$	$e^x$	$\sec x$	$\sec x \tan x$
$\ln x$	$\dfrac{1}{x}$	$\operatorname{cosec} x$	$-\operatorname{cosec} x \cot x$
$\sin x$	$\cos x$	$\cot x$	$-\operatorname{cosec}^2 x$
$\cos x$	$-\sin x$	$\tan^{-1} x$	$\dfrac{1}{1 + x^2}$

With a linear inside, the chain rule just multiplies by $a$ :

\frac{d}{dx} e^{ax+b} = a e^{ax+b}, \qquad \frac{d}{dx} \ln(ax+b) = \frac{a}{ax+b}, \qquad \frac{d}{dx}\sin(ax+b) = a\cos(ax+b)

The trigonometric results hold only when $x$ is in radians.

The list of formulae given in the exam contains most of the right-hand column and the product and quotient rules, but not the basics. More importantly, you will not have time to look things up while chaining several rules together. Know them cold.

Radians only

$\dfrac{d}{dx}\sin x = \cos x$ is true only when $x$ is measured in radians. If a question is set in degrees you must convert first. In calculus questions in P3, always set your calculator to radians and leave angles such as $\tfrac{\pi}{6}$ exact.

The chain rule, properly

The chain rule is the engine of P3 differentiation. It deals with a composite function, a function of a function, such as $\ln(3x^2 + 2)$ : first you compute $u = 3x^2 + 2$ (the inner function), then you take $\ln u$ (the outer function).

Why rates multiply

Suppose $u$ changes 6 times as fast as $x$ , and $y$ changes $\tfrac{1}{5}$ as fast as $u$ . Then $y$ changes $6 \times \tfrac{1}{5}$ times as fast as $x$ . Rates of change through a chain of dependencies multiply. That is all the chain rule says.

Chain rule

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 function notation: $\dfrac{d}{dx} f\big(g(x)\big) = f'\big(g(x)\big) \times g'(x)$ .

In words: differentiate the outside, keep the inside unchanged, then multiply by the derivative of the inside.

The form in words is the one to use in practice. For $\ln(3x^2 + 2)$ : the outside is $\ln(\ \cdot\ )$ , whose derivative is $\dfrac{1}{(\ \cdot\ )}$ ; keep the inside, giving $\dfrac{1}{3x^2+2}$ ; multiply by the derivative of the inside, $6x$ . Result: $\dfrac{6x}{3x^2+2}$ .

Layered composites

Some functions have three or more layers. Peel them from the outside in, one factor per layer.

$\sin^3 2x$ means $(\sin 2x)^3$ . The layers, from outside in, are: cube, sine, $2x$ .

\frac{d}{dx}(\sin 2x)^3 = \underbrace{3(\sin 2x)^2}_{\text{cube}} \times \underbrace{\cos 2x}_{\text{sine}} \times \underbrace{2}_{2x} = 6\sin^2 2x\cos 2x

Differentiating a composite

Write the function so its layers are visible: $\sin^3 2x = (\sin 2x)^3$ , $\sqrt{1 + e^x} = (1 + e^x)^{1/2}$ , $\dfrac{1}{\cos x} = (\cos x)^{-1}$ .
Differentiate the outermost layer, leaving everything inside it untouched.
Multiply by the derivative of the next layer in, and so on, until you reach $x$ .
Tidy the product: constants to the front, powers combined, a recognisable trig form if there is one.

The derivative of an inverse

A useful consequence: if you can write $x$ as a function of $y$ , then

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

This is how $\dfrac{d}{dx}\ln x$ and $\dfrac{d}{dx}\tan^{-1}x$ are found from $e^x$ and $\tan x$ . It also appears directly in questions such as "given $x = y^2 + \ln y$ , find $\dfrac{dy}{dx}$ in terms of $y$ ".

Reading the structure before you start

Most errors in P3 differentiation are not algebra slips; they are choosing the wrong rule. Before you write anything, ask what the last operation would be if you evaluated the expression for a particular $x$ .

Last operation	Example	Rule
Adding or subtracting	$e^{2x} - 3\ln x$	Differentiate term by term
Multiplying two functions of $x$	$x^2 e^{-x}$	Product rule
Dividing two functions of $x$	$\dfrac{\ln x}{x+1}$	Quotient rule
Applying a function to an expression	$\ln(\cos 2x)$ , $e^{x^2}$ , $(1 + \tan x)^4$	Chain rule
Multiplying by a constant	$5\sin 3x$	Keep the constant, differentiate the rest

Inside each rule, the pieces may themselves need the chain rule. For $x^2 e^{-x}$ the last operation is a multiplication (product rule), and differentiating $e^{-x}$ inside it needs the chain rule.

Simplify first

Some functions become much easier after a rewrite. Do it before differentiating, not after.

Logarithm laws. $\ln\sqrt{x^2+1} = \tfrac{1}{2}\ln(x^2 + 1)$ , and $\ln\dfrac{(2x+1)^3}{x-1} = 3\ln(2x+1) - \ln(x-1)$ . Differentiating the expanded form avoids a chain rule inside a quotient rule.
Indices. $\dfrac{3}{\sqrt{x}} = 3x^{-1/2}$ ; $\dfrac{e^{3x} + 1}{e^x} = e^{2x} + e^{-x}$ .
Splitting a fraction with a single-term denominator: $\dfrac{x^2 + 4}{x} = x + 4x^{-1}$ .
Trigonometric identities. $2\sin x\cos x = \sin 2x$ ; $\dfrac{\sin x}{\cos x} = \tan x$ .

Applications: gradients, tangents, normals and stationary points

The applications are the same as in P1. What makes them P3 questions is that the functions are harder and the answers are exact.

Using the derivative

Gradient at $x = a$ : $f'(a)$ .
Tangent at $(a, f(a))$ : $y - f(a) = f'(a)(x - a)$ .
Normal at $(a, f(a))$ : gradient $-\dfrac{1}{f'(a)}$ , since perpendicular gradients multiply to $-1$ .
Stationary point: where $\dfrac{dy}{dx} = 0$ .
Nature: $\dfrac{d^2y}{dx^2} > 0$ means a minimum, $\dfrac{d^2y}{dx^2} < 0$ a maximum. If it is $0$ or awkward, check the sign of $\dfrac{dy}{dx}$ either side.
Increasing where $\dfrac{dy}{dx} > 0$ , decreasing where $\dfrac{dy}{dx} < 0$ .

In P3, setting $\dfrac{dy}{dx} = 0$ often leads to an equation that cannot be solved exactly. Then the question says "show that the $x$ -coordinate of the stationary point satisfies the equation $x = \ldots$ " and continues with an iteration. See numerical methods for the second half.

Tip

Exponentials are never zero. When $\dfrac{dy}{dx}$ factorises as $e^{\text{something}} \times (\ldots)$ , the stationary points come only from the bracket. Say so in one line: "since $e^{-x} > 0$ , $\dfrac{dy}{dx} = 0 \Rightarrow 2x - x^2 = 0$ ."

Worked examples

Composites with each standard function

Differentiate with respect to $x$ :

(a) $e^{1 - 4x}$ (b) $\ln(3x^2 + 2)$ (c) $\sin^3 2x$ (d) $\tan^{-1} 2x$

Solution

(a) Outer $e^{(\cdot)}$ , inner $1 - 4x$ with derivative $-4$ :

\frac{d}{dx}e^{1-4x} = -4e^{1-4x}

(b) Outer $\ln(\cdot)$ , inner $3x^2 + 2$ with derivative $6x$ :

\frac{d}{dx}\ln(3x^2 + 2) = \frac{6x}{3x^2 + 2}

\frac{d}{dx}(\sin 2x)^3 = 3(\sin 2x)^2 \cdot \cos 2x \cdot 2 = 6\sin^2 2x \cos 2x

(d) Outer $\tan^{-1}(\cdot)$ with derivative $\dfrac{1}{1 + (\cdot)^2}$ , inner $2x$ :

\frac{d}{dx}\tan^{-1} 2x = \frac{1}{1 + (2x)^2} \times 2 = \frac{2}{1 + 4x^2}

Simplify, then differentiate

Find $\dfrac{dy}{dx}$ when (a) $y = \ln\sqrt{x^2 + 1}$ and (b) $y = \ln \dfrac{(2x + 1)^3}{x - 1}$ .

Solution

(a) Use $\ln a^k = k\ln a$ first: $y = \tfrac{1}{2}\ln(x^2 + 1)$ .

\frac{dy}{dx} = \frac{1}{2} \times \frac{2x}{x^2 + 1} = \frac{x}{x^2 + 1}

(b) Expand with the log laws: $y = 3\ln(2x + 1) - \ln(x - 1)$ .

\frac{dy}{dx} = \frac{3 \times 2}{2x + 1} - \frac{1}{x - 1} = \frac{6}{2x+1} - \frac{1}{x-1}

Over a common denominator this is $\dfrac{6(x-1) - (2x+1)}{(2x+1)(x-1)} = \dfrac{4x - 7}{(2x + 1)(x - 1)}$ .

Exact tangent

The curve $y = e^{2x} - 4x$ has a point $P$ where $x = \ln 2$ . Find the equation of the tangent at $P$ , and the exact $y$ -coordinate of the point where it crosses the $y$ -axis.

Solution

At $x = \ln 2$ : $e^{2\ln 2} = e^{\ln 4} = 4$ , so $y = 4 - 4\ln 2$ .

$\dfrac{dy}{dx} = 2e^{2x} - 4$ , so at $P$ the gradient is $2(4) - 4 = 4$ .

Tangent:

y - (4 - 4\ln 2) = 4(x - \ln 2) \quad\Rightarrow\quad y = 4x + 4 - 8\ln 2

At $x = 0$ the tangent has $y = 4 - 8\ln 2$ .

Stationary point with logarithms

Find the exact coordinates of the stationary point of $y = 3\ln(2x + 1) - 2x$ , for $x > -\tfrac{1}{2}$ , and determine its nature.

Solution

\frac{dy}{dx} = \frac{6}{2x + 1} - 2

Set equal to zero: $\dfrac{6}{2x+1} = 2 \Rightarrow 2x + 1 = 3 \Rightarrow x = 1$ .

Then $y = 3\ln 3 - 2$ . The stationary point is $(1,\ 3\ln 3 - 2)$ .

\frac{d^2y}{dx^2} = -\frac{12}{(2x + 1)^2}, \qquad \text{at } x = 1:\ -\frac{12}{9} = -\frac{4}{3} < 0

so it is a maximum.

Stationary points of a product

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

Solution

By the product rule,

\frac{dy}{dx} = 2xe^{-x} - x^2e^{-x} = e^{-x}(2x - x^2) = xe^{-x}(2 - x)

Since $e^{-x} > 0$ , $\dfrac{dy}{dx} = 0$ gives $x = 0$ or $x = 2$ . The points are $(0, 0)$ and $(2, 4e^{-2})$ .

Differentiating again (product rule on $e^{-x}(2x - x^2)$ ):

\frac{d^2y}{dx^2} = -e^{-x}(2x - x^2) + e^{-x}(2 - 2x) = e^{-x}(x^2 - 4x + 2)

At $x = 0$ : $2 > 0$ , a minimum. At $x = 2$ : $e^{-2}(4 - 8 + 2) = -2e^{-2} < 0$ , a maximum.

y = x^2 e^(-x) (2, 0) -- (2, 4 e^(-2))

The curve $y = x^2e^{-x}$ touches the origin (a minimum) and has a maximum at $x = 2$ , then decays towards the $x$ -axis.

Exam-hard: a stationary point you cannot solve exactly

The curve $y = \dfrac{\ln x}{x + 1}$ has one stationary point.

(a) Show that its $x$ -coordinate satisfies $x = e^{1 + 1/x}$ .

(b) Use the iteration $x_{n+1} = e^{1 + 1/x_n}$ with $x_1 = 3.5$ to find the $x$ -coordinate correct to 2 decimal places, showing the result of each iteration to 4 decimal places.

Solution

(a) Quotient rule with $u = \ln x$ , $v = x + 1$ :

\frac{dy}{dx} = \frac{(x+1)\cdot\frac{1}{x} - \ln x \cdot 1}{(x + 1)^2}

At a stationary point the numerator is zero:

\frac{x + 1}{x} = \ln x \quad\Rightarrow\quad \ln x = 1 + \frac{1}{x} \quad\Rightarrow\quad x = e^{1 + 1/x}

(b) $x_1 = 3.5$ , $x_2 = 3.6173$ , $x_3 = 3.5839$ , $x_4 = 3.5931$ , $x_5 = 3.5906$ , $x_6 = 3.5913$ , $x_7 = 3.5911$ .

The iterates agree to 2 d.p., so $x = 3.59$ (2 d.p.).

Common mistakes

Forgetting the inner derivative. $\dfrac{d}{dx}e^{x^2} = 2xe^{x^2}$ , not $e^{x^2}$ . $\dfrac{d}{dx}\sin 4x = 4\cos 4x$ , not $\cos 4x$ .
Treating $\ln(ax)$ like $\ln(ax+b)$ wrongly. $\dfrac{d}{dx}\ln 7x = \dfrac{7}{7x} = \dfrac{1}{x}$ . That is correct, because $\ln 7x = \ln 7 + \ln x$ .
Misreading powers of trig functions. $\sin^2 x$ means $(\sin x)^2$ , derivative $2\sin x\cos x$ . It is not $\sin(x^2)$ , whose derivative is $2x\cos(x^2)$ .
Normal gradient. The normal gradient is $-\dfrac{1}{m}$ , not $-m$ and not $\dfrac{1}{m}$ .
Using degrees. Calculator in degree mode when substituting $x = 0.5$ gives nonsense.
Dividing by a possible zero. When solving $x e^{-x}(2 - x) = 0$ , do not divide by $x$ ; you lose the root $x = 0$ .

Exam tip

"Find the exact coordinates" or "exact value" means no decimals anywhere: leave $\ln 3$ , $e^{-2}$ , $\sqrt{3}$ , $\pi$ in the answer. A correct decimal usually loses the final mark.
"Show that" questions give you the answer. Every step must be visible, including setting the derivative to zero and the rearrangement. The final line should match the printed result exactly.
For the nature of a stationary point, state the sign of the second derivative and the conclusion: " $\dfrac{d^2y}{dx^2} = -\tfrac{4}{3} < 0$ , so maximum." The conclusion without the evidence earns nothing.
When differentiation is the first part of a longer question (iteration, area, equation of a tangent), a wrong derivative costs you the whole question. Check it by substituting a value of $x$ and comparing with a quick numerical gradient on your calculator, if time allows.

Summary

The meaning of a derivative is unchanged; P3 adds $e^x$ , $\ln x$ , all six trig functions and $\tan^{-1}x$ .
The chain rule: differentiate the outside, keep the inside, multiply by the derivative of the inside. Peel layered composites one layer at a time.
Linear insides just multiply by $a$ : $\dfrac{d}{dx}e^{ax+b} = ae^{ax+b}$ , $\dfrac{d}{dx}\ln(ax+b) = \dfrac{a}{ax+b}$ .
Identify the last operation to choose the rule: sum, product, quotient or composite.
Simplify first with log laws, indices and identities.
$\dfrac{dy}{dx} = 1 \Big/ \dfrac{dx}{dy}$ .
Trig derivatives need radians.
Tangent $y - y_1 = m(x - x_1)$ ; normal gradient $-1/m$ ; stationary points where $\dfrac{dy}{dx} = 0$ ; nature from the sign of $\dfrac{d^2y}{dx^2}$ .
Exact answers stay exact.

Practice

Question

Differentiate (a) $4e^{-x/2}$ (b) $\ln(5 - 2x)$ (c) $\cos^2 3x$ (d) $\tan^{-1}\dfrac{x}{3}$ .
Given $y = \ln\big(x^3\sqrt{2x - 1}\big)$ , find $\dfrac{dy}{dx}$ .
Find the exact gradient of $y = e^{2x}\sin x$ at the point where $x = \tfrac{\pi}{4}$ .
Find the equation of the normal to $y = \ln(3x - 2)$ at the point where $x = 1$ , and the point where it meets the $y$ -axis.
Find the exact coordinates of the stationary point of $y = xe^{-2x}$ and determine its nature.
Given $x = y^2 + \ln y$ for $y > 0$ , find $\dfrac{dy}{dx}$ in terms of $y$ , and its value when $y = 1$ .
The curve $y = e^{2x} - 6e^x + 4x$ has two stationary points. Find their exact coordinates and determine the nature of each.
Find the exact $x$ -coordinates of the stationary points of $y = (x + 1)e^{-x^2}$ , and state, with a reason, which is a maximum.
The curve $y = \dfrac{e^x}{x^2 + 1}$ is defined for all $x$ . Show that $\dfrac{dy}{dx} \geq 0$ for all $x$ , and find the coordinates of the point where the gradient is zero.

Answers

(a) $4 \times \left(-\tfrac{1}{2}\right)e^{-x/2} = -2e^{-x/2}$ . (b) $\dfrac{-2}{5 - 2x}$ . (c) $2\cos 3x \times (-\sin 3x) \times 3 = -6\sin 3x\cos 3x$ , which is $-3\sin 6x$ . (d) $\dfrac{1}{1 + x^2/9} \times \dfrac{1}{3} = \dfrac{3}{9 + x^2}$ .
Simplify: $y = 3\ln x + \tfrac{1}{2}\ln(2x - 1)$ . Then $\dfrac{dy}{dx} = \dfrac{3}{x} + \dfrac{1}{2} \cdot \dfrac{2}{2x - 1} = \dfrac{3}{x} + \dfrac{1}{2x - 1}$ .
$\dfrac{dy}{dx} = 2e^{2x}\sin x + e^{2x}\cos x = e^{2x}(2\sin x + \cos x)$ . At $x = \tfrac{\pi}{4}$ : $e^{\pi/2}\left(\tfrac{2}{\sqrt{2}} + \tfrac{1}{\sqrt{2}}\right) = \dfrac{3}{\sqrt{2}}e^{\pi/2} = \dfrac{3\sqrt{2}}{2}e^{\pi/2}$ .
At $x = 1$ , $y = \ln 1 = 0$ . $\dfrac{dy}{dx} = \dfrac{3}{3x - 2} = 3$ at $x = 1$ , so the normal gradient is $-\tfrac{1}{3}$ . Normal: $y = -\tfrac{1}{3}(x - 1)$ , i.e. $x + 3y = 1$ . It meets the $y$ -axis at $\left(0, \tfrac{1}{3}\right)$ .
$\dfrac{dy}{dx} = e^{-2x} - 2xe^{-2x} = e^{-2x}(1 - 2x) = 0 \Rightarrow x = \tfrac{1}{2}$ , $y = \tfrac{1}{2}e^{-1}$ . $\dfrac{d^2y}{dx^2} = e^{-2x}(4x - 4)$ , which at $x = \tfrac{1}{2}$ is $-2e^{-1} < 0$ : a maximum at $\left(\tfrac{1}{2}, \tfrac{1}{2e}\right)$ .
$\dfrac{dx}{dy} = 2y + \dfrac{1}{y} = \dfrac{2y^2 + 1}{y}$ , so $\dfrac{dy}{dx} = \dfrac{y}{2y^2 + 1}$ . At $y = 1$ : $\tfrac{1}{3}$ .
$\dfrac{dy}{dx} = 2e^{2x} - 6e^x + 4 = 2(e^x - 1)(e^x - 2)$ . Zero when $e^x = 1$ or $e^x = 2$ : $x = 0$ or $x = \ln 2$ . Points $(0, -5)$ and $(\ln 2,\ 4\ln 2 - 8)$ . $\dfrac{d^2y}{dx^2} = 4e^{2x} - 6e^x$ : at $x = 0$ it is $-2 < 0$ (maximum); at $x = \ln 2$ it is $16 - 12 = 4 > 0$ (minimum).
$\dfrac{dy}{dx} = e^{-x^2} - 2x(x + 1)e^{-x^2} = e^{-x^2}(1 - 2x - 2x^2)$ . Since $e^{-x^2} > 0$ , solve $2x^2 + 2x - 1 = 0$ : $x = \dfrac{-1 \pm \sqrt{3}}{2}$ . The quadratic $1 - 2x - 2x^2$ is positive between its roots and negative outside, so $\dfrac{dy}{dx}$ goes $-, +, -$ . Hence $x = \dfrac{-1 - \sqrt{3}}{2}$ is a minimum and $x = \dfrac{-1 + \sqrt{3}}{2}$ is a maximum (gradient changes from positive to negative).
Quotient rule: $\dfrac{dy}{dx} = \dfrac{(x^2 + 1)e^x - e^x \cdot 2x}{(x^2 + 1)^2} = \dfrac{e^x(x - 1)^2}{(x^2 + 1)^2}$ . Every factor is non-negative, so $\dfrac{dy}{dx} \geq 0$ for all $x$ . It is zero only at $x = 1$ , giving the point $\left(1, \tfrac{e}{2}\right)$ . (This is a stationary point of inflexion: the gradient is positive on both sides.)

What changes from P1

The standard derivatives

The chain rule, properly

Why rates multiply

Layered composites

The derivative of an inverse

Reading the structure before you start

Simplify first

Applications: gradients, tangents, normals and stationary points

Worked examples

Practice

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