2. Rules & Standard Derivatives

Video coming soon4 worked examples

Theory

Advanced Higher extends Higher differentiation with three key rules and a wider set of standard derivatives.

(uv)' = u'v + uv' \qquad \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \qquad \bigl(f(g(x))\bigr)' = f'(g(x))\,g'(x)

It also introduces three reciprocal trigonometric functions, which you must know by name:

\sec\theta = \frac{1}{\cos\theta}, \qquad \operatorname{cosec}\theta = \frac{1}{\sin\theta}, \qquad \cot\theta = \frac{1}{\tan\theta} = \frac{\cos\theta}{\sin\theta}

Note the pairing: $\sec$ goes with $\cos$ , and $\operatorname{cosec}$ goes with $\sin$ — not the other way round. Together with the exponential and logarithmic functions, these give the standard derivatives for this course:

\begin{aligned} \frac{d}{dx}(\tan x) &= \sec^2 x & \frac{d}{dx}(\cot x) &= -\operatorname{cosec}^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}(e^x) &= e^x & \frac{d}{dx}(\ln x) &= \frac{1}{x} \end{aligned}

The two functions beginning with “co” — $\cot$ and $\operatorname{cosec}$ — are the ones with negative derivatives, exactly as $\cos$ is. (The inverse trigonometric functions $\sin^{-1}$ , $\cos^{-1}$ and $\tan^{-1}$ have their own topic.)

The Golden Rule: identify the structure before differentiating — a product, a quotient, or a composition — and apply the matching rule. For combinations, work from the outside in.

⚠️ Common Examiner Traps

Quotient rule order: the numerator is $u'v - uv'$ , in that order — reversing it flips the sign.
Chain rule inside factor: always multiply by the derivative of the inner function.
Standard derivatives: know the exact forms, e.g. $\frac{d}{dx}(\tan x) = \sec^2 x$ , not $\sec x$ .
Mixing up sec and cosec: $\sec\theta = \frac{1}{\cos\theta}$ — the names are deliberately crossed over, and it is a costly slip.
Losing the minus sign: $\cot x$ and $\operatorname{cosec} x$ both differentiate to negative expressions.
Powers of trig functions: $\sec^3 x$ means $(\sec x)^3$ , so it needs the chain rule — differentiate the power first, then multiply by $\sec x \tan x$ .

Worked examples

Example 1

Differentiate $y = x^2 \sin x$ .

Step 1: This is a product with $u = x^2$ , $v = \sin x$ . Apply the product rule:

\frac{dy}{dx} = (2x)(\sin x) + (x^2)(\cos x)

Step 2: So $\dfrac{dy}{dx} = 2x\sin x + x^2\cos x$ .

Example 2

Differentiate $y = \dfrac{2x + 1}{x^2 + 1}$ .

Step 1: Quotient rule with $u = 2x+1$ ( $u' = 2$ ) and $v = x^2+1$ ( $v' = 2x$ ):

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

Step 2: Expand and simplify the numerator:

= \frac{2x^2 + 2 - 4x^2 - 2x}{(x^2+1)^2} = \frac{-2x^2 - 2x + 2}{(x^2+1)^2}

Example 3

Differentiate $y = e^{3x}\cos 2x$ .

Step 1: Product rule, using the chain rule on each factor: $\frac{d}{dx}(e^{3x}) = 3e^{3x}$ and $\frac{d}{dx}(\cos 2x) = -2\sin 2x$ .

\frac{dy}{dx} = 3e^{3x}\cos 2x + e^{3x}(-2\sin 2x)

Step 2: Factor out $e^{3x}$ :

= e^{3x}(3\cos 2x - 2\sin 2x)

Example 4

Differentiate (a) $y = 5\operatorname{cosec} 3x$ , (b) $y = \sec^3 x$ , and (c) $y = x\cot x$ .

Step 1 (a): Use $\frac{d}{dx}(\operatorname{cosec} x) = -\operatorname{cosec} x \cot x$ together with the chain rule, since the angle is $3x$ :

\frac{dy}{dx} = 5 \times \bigl(-\operatorname{cosec} 3x \cot 3x\bigr) \times 3 = -15\operatorname{cosec} 3x \cot 3x

Step 2 (b): Read $\sec^3 x$ as $(\sec x)^3$ . Differentiate the power first, then multiply by the derivative of $\sec x$ :

\frac{dy}{dx} = 3(\sec x)^2 \times \sec x \tan x = 3\sec^3 x \tan x

Step 3 (c): This is a product, so use $u'v + uv'$ with $u = x$ and $v = \cot x$ , remembering the minus sign:

\frac{dy}{dx} = (1)\cot x + x\bigl(-\operatorname{cosec}^2 x\bigr)

Step 4: Tidy the result:

\frac{dy}{dx} = \cot x - x\operatorname{cosec}^2 x