Differentiating composite functions with the chain rule, including expressions of the form a function of a linear expression and sine and cosine of a linear expression, and reversing the process to integrate functions of the form (ax + b) to the n, sin(ax + b) and cos(ax + b).

What this dot point is asking

The SQA wants you to differentiate a composite function, one function nested inside another, using the chain rule, and then to reverse the process to integrate functions of the form $(ax + b)^n$, $\sin(ax + b)$ and $\cos(ax + b)$. At Higher the inside function is almost always a simple expression such as $3x + 2$ or $2x^3 + 1$, so the chain rule reduces to a quick "differentiate the outside, multiply by the inside derivative" routine, and integration reverses it by dividing by that inside derivative.

The chain rule

A composite function is built by feeding one function into another. Writing $y = f(g(x))$, the inside function $g$ acts first and the outside function $f$ acts on the result. The power rule alone cannot differentiate $(2x + 5)^7$ correctly, because the variable is wrapped inside a bracket; the chain rule supplies the missing factor.

The single most reliable way to apply it is to name the outside and the inside before you start, differentiate the outside, then tack on the inside derivative as a multiplier.

Trigonometric composites

The chain rule handles $\sin$ and $\cos$ of a linear expression in exactly the same way. The outside derivatives are the standard ones, $\dfrac{d}{dx}\sin x = \cos x$ and $\dfrac{d}{dx}\cos x = -\sin x$ (in radians), and the inside derivative is the multiplier.

Reversing the chain rule to integrate

Because integration undoes differentiation, the chain rule run backwards tells you how to integrate a bracket raised to a power, or a sine or cosine of a linear expression. The key idea is that differentiation multiplies by the inside derivative $a$, so integration must divide by it.

Examples in context

When a model gives a quantity as a function of a scaled or shifted variable, the chain rule is what lets you find its rate of change. If the height of a tide is modelled by $h(t) = 3\sin(0.5t + 1)$ metres, then the rate at which the height changes is $h'(t) = 3 \cdot 0.5\cos(0.5t + 1) = 1.5\cos(0.5t + 1)$ metres per unit time, where the factor $0.5$ comes straight from the chain rule. Reversing the process, an engineer who knows a velocity of the form $v(t) = (2t + 1)^3$ recovers displacement by integrating, dividing by the inside derivative $2$ to get $\dfrac{(2t + 1)^4}{8} + c$.

Try this

Q1. Differentiate $y = (4x + 3)^6$. [3 marks]

• Cue. Outside derivative $6(4x + 3)^5$, inside derivative $4$, so $\dfrac{dy}{dx} = 24(4x + 3)^5$.

Q2. Find $\displaystyle\int \cos(2x + 5)\,dx$. [2 marks]

• Cue. $\dfrac{1}{2}\sin(2x + 5) + c$, dividing by the inside derivative $2$ and remembering the constant.