The chain, product and quotient rules, implicit differentiation, derivatives of exponential, logarithmic and trigonometric functions, and the second derivative and concavity.

What this dot point is asking

WJEC wants you to differentiate using the chain, product and quotient rules, to use implicit differentiation for relations not solved for $y$, to know the derivatives of the standard functions ($\mathrm{e}^x$, $\ln x$, $\sin x$, $\cos x$, $\tan x$), and to use the second derivative for concavity and to classify stationary points. This extends the AS power-rule work to the full range of functions.

The answer

The chain, product and quotient rules

The chain rule differentiates a function of a function: for $y = (3x + 1)^5$, take $u = 3x + 1$, so $\dfrac{dy}{dx} = 5(3x+1)^4 \times 3 = 15(3x+1)^4$.

Standard derivatives

Implicit differentiation

When a relation links $x$ and $y$ without $y$ being the subject (such as $x^2 + y^2 = 25$), differentiate both sides with respect to $x$, treating $y$ as a function of $x$ and applying the chain rule to any $y$ terms.

The second derivative and concavity

The second derivative $\dfrac{d^2y}{dx^2}$ measures how the gradient is changing. It classifies stationary points: positive means a minimum (concave up), negative means a maximum (concave down). A point of inflection is where concavity changes, often where $\dfrac{d^2y}{dx^2} = 0$ and changes sign.

Examples in context

Example 1. Connected rates of change. A spherical balloon has volume $V = \tfrac{4}{3}\pi r^3$ and is inflated so that $\dfrac{dV}{dt} = 50\,\text{cm}^3\text{s}^{-1}$. By the chain rule $\dfrac{dV}{dt} = \dfrac{dV}{dr}\dfrac{dr}{dt} = 4\pi r^2 \dfrac{dr}{dt}$, so when $r = 5$, $\dfrac{dr}{dt} = \dfrac{50}{4\pi(25)} \approx 0.159\,\text{cm s}^{-1}$. The chain rule connects the rate of volume change to the rate of radius change.

Example 2. A product turning point. For $y = x\mathrm{e}^{-x}$, the product rule gives $\dfrac{dy}{dx} = \mathrm{e}^{-x}(1 - x)$. Setting this to zero gives $x = 1$, and the second derivative there is negative, so $(1, \mathrm{e}^{-1})$ is a maximum. Product-rule differentiation locates and classifies the peak.

Try this

Q1. Differentiate $y = (2x + 3)^4$. [2 marks]

• Cue. Chain rule: $\dfrac{dy}{dx} = 4(2x+3)^3 \times 2 = 8(2x+3)^3$.

Q2. Differentiate $y = x\ln x$. [2 marks]

• Cue. Product rule: $\dfrac{dy}{dx} = x \cdot \dfrac{1}{x} + \ln x \cdot 1 = 1 + \ln x$.

Q3. For $x^2 + xy = 6$, find $\dfrac{dy}{dx}$. [3 marks]

• Cue. Differentiate: $2x + \left(y + x\dfrac{dy}{dx}\right) = 0$, so $\dfrac{dy}{dx} = -\dfrac{2x + y}{x}$.