Chain, product and quotient rules · Calculus (JAMB bridge)

Sine and cosine

With $x$ in radians, the gradient of $y = \sin x$ at every point is $\cos x$ , and the gradient of $y = \cos x$ is $-\sin x$ . The board below shows it: the gradient graph of $\sin x$ is exactly the curve $\cos x$ .

The gradient of a curveSlide P along the curve

(0.5, 0.48)P0.88gradient of tangent

x at P = 0.5

x is in radians. The derivative is dy/dx = cos x, so at x = 0.5 the tangent's gradient is 0.88: the curve is going up. The grey dashed curve plots the gradient at every x: it is the graph of y = cos x, and it crosses zero where the tangent is flat.

\begin{aligned} \frac{d}{dx}\sin x &= \cos x \\ \frac{d}{dx}\cos x &= -\sin x \end{aligned}

The cycleEach differentiation moves one step round: sin → cos → −sin → −cos → sin

The chain rule

For a function inside another function, such as $(3x + 1)^5$ or $\sin(4x)$ , differentiate the outside (leaving the inside alone), then multiply by the derivative of the inside:

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

where $u$ is the inside.

y = (3x + 1)⁵

dy/dx = 5(3x + 1)⁴ × 3

differentiate the outside, leaving the inside alone, then × the inside’s derivative

The chain ruleOutside first, then times the derivative of the inside

The product rule

For a product of two functions, $y = uv$ :

\frac{dy}{dx} = u'v + uv'

Differentiate one factor at a time, keeping the other as it is, and add. The picture shows why: when $u$ and $v$ both grow a little, the area $uv$ grows by two strips.

The product ruleThe rectangle's area uv grows by v·δu + u·δv

The quotient rule

For $y = \dfrac uv$ :

\frac{dy}{dx} = \frac{vu' - uv'}{v^2}

The order matters because of the minus sign: bottom times the top’s derivative first.

d/dx (u ⁄ v) = (v·u′ − u·v′) ⁄ v²

bottom × derivative of top, minus top × derivative of bottom, all over bottom squared

The quotient ruleBottom × top′ − top × bottom′, over bottom squared

Try it

Chain, product and quotient rulesPress Next

y = (5x − 2)³

dy/dx = (derivative of the outside) × (derivative of the inside)

The parts……

Differentiate each……

Multiply……

dy/dx = …

Chain rulerule0 of 4step

y = (5x − 2)³ needs the chain rule: dy/dx = (derivative of the outside) × (derivative of the inside). Press Next.

Step through each example. Watch how the two coloured parts move through the rule.

The second derivative

Differentiating $\frac{dy}{dx}$ again gives the second derivative, $\dfrac{d^2y}{dx^2}$ . It is used in the next lesson to tell maximum points from minimum points.

Worked example · NECO 2023

NECO 2023 · Paper 2 · Q4 (b)

Differentiate $(2x^2 + 5)^4$ with respect to $x$ .

Outside and inside
The outside is $(\;)^4$ and the inside is $u = 2x^2 + 5$ .
Think first. What is the inside of (2x² + 5)⁴?
Differentiate each
The outside gives $4(2x^2 + 5)^3$ ; the inside gives $4x$ .
Think first. Outside, then inside.
Multiply
$\frac{dy}{dx} = 16x(2x^2 + 5)^3$ .
Think first. 4(2x² + 5)³ × 4x = ?

Worked example · JAMB 1999

JAMB 1999 · UME · Q39

What is the derivative of $t^2\sin(3t - 5)$ with respect to $t$ ?

Which rule?
It is a product: $u = t^2$ and $v = \sin(3t - 5)$ .
Think first. What is multiplied by what?
Differentiate each
$u' = 2t$ and $v' = 3\cos(3t - 5)$ .
Think first. v needs the chain rule.
Product rule
$2t\sin(3t - 5) + 3t^2\cos(3t - 5)$ : option C.
Think first. u′v + uv′.

Your turn

NECO 2024 · Paper 2 · Q5 (a)

(a)
If $y = (3x^3 + 2x^2 + 1)(3x^2 + 4)$ , find $\dfrac{dy}{dx}$ .

Worked solution (try it first)

(a)

Use the product rule: if $y = uv$ , then $\frac{dy}{dx} = v\frac{du}{dx} + u\frac{dv}{dx}$ .
Here $u = 3x^3 + 2x^2 + 1$ , so $\frac{du}{dx} = 9x^2 + 4x$ , and $v = 3x^2 + 4$ , so $\frac{dv}{dx} = 6x$ .
Then $\frac{dy}{dx} = (9x^2 + 4x)(3x^2 + 4) + (3x^3 + 2x^2 + 1)(6x)$ .
Expand: $(27x^4 + 12x^3 + 36x^2 + 16x) + (18x^4 + 12x^3 + 6x)$ .
Collect like terms: $\frac{dy}{dx} = 45x^4 + 24x^3 + 36x^2 + 22x$ .

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