Derivatives of Powers, eˣ, ln x and Trig

Understand dy/dx as the limit of a chord's gradient, and differentiate any power of x, e to the ax, sin ax, cos ax, and (0606/4037) tan x and ln x.

In the exam for IGCSE Additional Maths 0606 · 1 real past-paper questions below

اقرأ هذا الدرس مشروحاً بالعربي ←

What dy/dx really means

You can already differentiate polynomials with the power rule. In this lesson you add new curves: $e^{x}$ , the trig graphs and, for 0606 and 4037, $ln x$ . First, a clear picture of what $\frac{d y}{d x}$ means.

Take a point on a curve. Move a tiny step along to a second point. The straight line joining them is a chord. Its gradient is the rise over the run.

Formula

\frac{d y}{d x} = δ x \to 0 lim \frac{δ y}{δ x}

$δ x$ — a small change in x (read it “delta x”). It is one symbol, not δ times x.
$δ y$ — the change in y that goes with it.
$δ x \to 0$ — “delta x tends to zero”: make the step smaller and smaller.
$\frac{d y}{d x}$ — what the chord gradient settles on. It is the gradient of the tangent.

Worked example

1
Watch it happen on $y = x^{2}$ , from $x$ to $x + δ x$ .
2
$δ y = (x + δ x)^{2} - x^{2} = 2 x δ x + (δ x)^{2}$ — The change in y: new height minus old height.
3
$\frac{δ y}{δ x} = 2 x + δ x$ — Divide by the run, δx. This is the chord's gradient.
4
$\frac{d y}{d x} = 2 x$ — Now let δx shrink to 0. The extra δx disappears. The power rule gives the same answer.

Rule

You need the idea, not the algebra. Neither board asks you to differentiate from first principles like this.

The answer is a new function that gives the gradient at every x. It is called the derived function, or the derivative. You will see it written in more than one way.

Rule

$\frac{d y}{d x}$ and $f^{'} (x)$ — the first derivative. They mean the same thing.
$\frac{d ^{2} y}{d x ^{2}}$ , $\frac{d}{d x} (\frac{d y}{d x})$ and $f^{''} (x)$ — the second derivative: differentiate twice.

Key idea

Newton and Leibniz both invented calculus but wrote it differently. Leibniz wrote dy/dx. Newton put a dot over the letter, and physicists still use his dot for rates over time.

Any power of x

The power rule works for every rational power: whole numbers, negatives and fractions. The only skill is to write each term as a number times a power of x first.

Formula

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

$n$ — the power of x. It can be negative or a fraction.
$a$ — the number in front. It waits and multiplies the answer.

Rule

Learn it. 0606 and 4037 give no calculus formulas. 4PM1 lists it with the formulas to learn.

Remember the rewrites: $x = x^{\frac{1}{2}}$ , $3 x = x^{\frac{1}{3}}$ , and x on the bottom gets a negative power, $\frac{1}{x ^{3}} = x^{- 3}$ .

Worked example

1
Find $\frac{d y}{d x}$ for $y = 6 3 x - \frac{4}{x} + \frac{2}{x ^{3}}$ , and the gradient at $x = 1$ .
2
$y = 6 x^{\frac{1}{3}} - 4 x^{- \frac{1}{2}} + 2 x^{- 3}$ — Rewrite every term as a power of x.
3
$6 x^{\frac{1}{3}} \to 6 \times \frac{1}{3} x^{- \frac{2}{3}} = 2 x^{- \frac{2}{3}}$ — Bring down 1/3. Then 1/3 − 1 = −2/3.
4
$- 4 x^{- \frac{1}{2}} \to - 4 \times (- \frac{1}{2}) x^{- \frac{3}{2}} = 2 x^{- \frac{3}{2}}$ — Bring down −1/2. Then −1/2 − 1 = −3/2. Two negatives make a positive.
5
$2 x^{- 3} \to - 6 x^{- 4}$ — Bring down −3. Then −3 − 1 = −4.
6
$\frac{d y}{d x} = 2 x^{- \frac{2}{3}} + 2 x^{- \frac{3}{2}} - 6 x^{- 4}$ — Put the terms together.
7
$x = 1 : 2 + 2 - 6 = - 2$ — 1 to any power is 1, so no calculator is needed.

Worked example

1
A fraction with a sum on top: $y = \frac{x ^{2} - 3}{x}$ .
2
$y = x^{\frac{3}{2}} - 3 x^{- \frac{1}{2}}$ — Split the fraction. Dividing by $x$ , which is $x^{\frac{1}{2}}$ , takes 1/2 off each power.
3
$\frac{d y}{d x} = \frac{3}{2} x^{\frac{1}{2}} + \frac{3}{2} x^{- \frac{3}{2}}$ — Power rule on each term.

Common mistake

Taking 1 off a negative power the wrong way. $- \frac{1}{2} - 1 = - \frac{3}{2}$ , not $- \frac{1}{2}$ or $\frac{1}{2}$ . The power always moves further from zero.

Key idea

Rewrite first, then differentiate. Every power of x follows the same rule.

e to the x: its own gradient

The number $e$ is about 2.718. The curve $y = e^{x}$ has a special property: at every point, its gradient is equal to its height. So differentiating it changes nothing.

Formula

\frac{d}{d x} (e^{x}) = e^{x}

When the power is a number times x, that number comes down in front. The power itself stays the same.

Formula

\frac{d}{d x} (e^{a x}) = a e^{a x}

$a$ — the number multiplying x in the power. For $e^{- x}$ , a = −1.

Rule

Learn both. 4PM1 lists $e^{a x}$ with the formulas to learn. 0606 and 4037 give none.

Worked example

1
$y = 5 e^{2 x} - 3 e^{- x} + x^{2}$ . Find the gradient at $x = 0$ .
2
$5 e^{2 x} \to 5 \times 2 e^{2 x} = 10 e^{2 x}$ — a = 2 comes down. The power 2x does not change.
3
$- 3 e^{- x} \to - 3 \times (- 1) e^{- x} = 3 e^{- x}$ — a = −1. Minus times minus is plus.
4
$x^{2} \to 2 x$ — The power rule, as before.
5
$\frac{d y}{d x} = 10 e^{2 x} + 3 e^{- x} + 2 x$ — Put the terms together.
6
$x = 0 : 10 (1) + 3 (1) + 0 = 13$ — Any number to the power 0 is 1, so $e^{0} = 1$ .

Common mistake

Using the power rule on $e^{x}$ . The x is in the power, not the base, so $e^{2 x}$ does not become $2 x e^{2 x - 1}$ .

Common mistake

Losing the minus from $e^{- x}$ . Its derivative is $- e^{- x}$ .

Key idea

On 0606 and 4037 Paper 1 there is no calculator. Leave answers exact unless the question asks for decimals. For example, the gradient above at $x = 1$ is $10 e^{2} + 3 e^{- 1} + 2$ .

sin and cos

The sine and cosine graphs are waves. Where $sin x$ is steepest going up, $cos x$ is at its top. That is no accident: the gradient of $sin x$ is $cos x$ .

Rule

These rules are only true when x is in radians. In calculus questions, angles are always in radians. Set your calculator to radians.

Formula

\frac{d}{d x} (sin a x) = a cos a x

$a$ — the number multiplying x. When a = 1: sin x becomes cos x.

Formula

\frac{d}{d x} (cos a x) = - a sin a x

$a$ — the same number. When a = 1: cos x becomes −sin x. Note the minus.

Rule

Learn both. 4PM1 lists them with the formulas to learn. 0606 and 4037 give no calculus formulas.

Memory trick

Go round the circle: sin → cos → −sin → −cos → back to sin. Each arrow is one differentiation.

Worked example

1
$y = 4 sin 3 x + 6 cos 2 x$ . Find the exact gradient at $x = \frac{π}{6}$ , then find $\frac{d ^{2} y}{d x ^{2}}$ .
2
$4 sin 3 x \to 4 \times 3 cos 3 x = 12 cos 3 x$ — a = 3 comes down. Sin becomes cos.
3
$6 cos 2 x \to 6 \times (- 2 sin 2 x) = - 12 sin 2 x$ — a = 2 comes down. Cos becomes minus sin.
4
$\frac{d y}{d x} = 12 cos 3 x - 12 sin 2 x$ — Put the terms together.
5
$x = \frac{π}{6} : 12 cos \frac{π}{2} - 12 sin \frac{π}{3}$ — Substitute. 3 × π/6 = π/2 and 2 × π/6 = π/3.
6
$= 12 (0) - 12 (\frac{3}{2}) = - 63$ — Use exact values. No calculator needed.
7
$\frac{d ^{2} y}{d x ^{2}} = - 36 sin 3 x - 24 cos 2 x$ — Differentiate again: cos 3x → −3 sin 3x, and −sin 2x → −2 cos 2x.

Common mistake

Getting the sign wrong: $cos x$ differentiates to $- sin x$ , but $sin x$ differentiates to $+ cos x$ .

Common mistake

Forgetting the a. $sin 3 x$ gives $3 cos 3 x$ , not $cos 3 x$ .

Key idea

Sin and cos swap places when you differentiate. Only cos picks up a minus.

tan x and ln x (0606/4037)

Rule

This step is for 0606 and 4037 only. 4PM1 does not have $tan x$ or $ln x$ in its list of derivatives to learn.

Two more standard functions. Learn them; no formula list gives them.

Formula

\frac{d}{d x} (tan x) = s e c^{2} x

$sec^{2} x$ — means $\frac{1}{c o s ^{2} x}$ . It is never negative, so tan x is always going uphill.

Formula

\frac{d}{d x} (ln x) = \frac{1}{x}

$x$ — must be positive, because ln only works on positive numbers.
$ln$ — the natural log, log to base e.

Use the log laws before you differentiate. They turn hard-looking terms into easy ones.

Example

$ln 4 x = ln 4 + ln x$ . The $ln 4$ is a plain number, so its derivative is 0. So $\frac{d}{d x} (ln 4 x) = \frac{1}{x}$ .

Worked example

1
$y = 2 tan x + ln 4 x + ln (x^{3})$ . Find $\frac{d y}{d x}$ , and the exact gradient at $x = \frac{π}{4}$ .
2
$2 tan x \to 2 sec^{2} x$ — Standard result, times 2.
3
$ln 4 x = ln 4 + ln x \to \frac{1}{x}$ — Split with the log law. ln 4 is a constant.
4
$ln (x^{3}) = 3 ln x \to \frac{3}{x}$ — The power comes to the front as a multiplier first.
5
$\frac{d y}{d x} = 2 sec^{2} x + \frac{4}{x}$ — Put the terms together: 1/x + 3/x = 4/x.
6
$sec^{2} \frac{π}{4} = \frac{1}{( \frac{1}{2} ) ^{2}} = 2$ — cos π/4 is 1 over root 2. Square it and flip it.
7
$x = \frac{π}{4} : 2 (2) + \frac{4}{\frac{π}{4}} = 4 + \frac{16}{π}$ — Exact answer. Leave π in it.

Common mistake

Writing $\frac{d}{d x} (ln 4 x) = \frac{1}{4 x}$ . The answer is $\frac{1}{x}$ : the 4 only adds a constant, ln 4.

Common mistake

Writing $\frac{d}{d x} (ln x^{3}) = \frac{1}{x ^{3}}$ . Bring the power down first: $3 ln x$ gives $\frac{3}{x}$ .

Key idea

With ln, tidy with the log laws first. Then each piece is just 1 over x.

Which rule? Read the question

Look at each term. Its shape tells you the rule.

You see	You do
A root, or x on the bottom	Rewrite as $x^{n}$ , then bring down n and take 1 off
$e^{a x}$	$a e^{a x}$ : the power stays the same
$sin a x$ or $cos a x$	$a cos a x$ or $- a sin a x$ , with x in radians
$tan x$ (0606 / 4037)	$sec^{2} x$
$ln a x$ or $ln x^{k}$ (0606 / 4037)	Log laws first: $\frac{1}{x}$ or $\frac{k}{x}$
$f^{''} (x)$ or $\frac{d ^{2} y}{d x ^{2}}$	Differentiate, then differentiate again
“Exact”, or 0606 / 4037 Paper 1	Keep e, π, surds and ln in the answer

A bracket with a power, like $(2 x + 1)^{5}$ , or two functions multiplied together, needs more: see the chain rule and product rule lessons.

Common mistake

Mixing up the e and power rules. $x^{3}$ uses the power rule. $e^{3 x}$ keeps its power and gains a 3 in front.

Key idea

Write every term in a standard shape first. Then each one has its own one-line rule.

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Real past-paper questions on derivatives of powers, eˣ, ln x and trig

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0606/23 May/Jun 2025 Q2 · 3 marks

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