Implicit differentiation and second derivatives: VCE Specialist Mathematics Unit 3

What this dot point is asking

VCAA wants you to differentiate relations given implicitly, such as $x^2 + y^2 = 25$, where $y$ is not isolated, and to use the second derivative to determine concavity and locate points of inflection. These tools extend the differentiation of Methods to curves that are not functions and deepen curve sketching.

Implicit differentiation

When a curve is given by an equation linking $x$ and $y$, such as $x^2 + y^2 = 25$, we can find $\frac{dy}{dx}$ without solving for $y$. Differentiate every term with respect to $x$, remembering that $y$ depends on $x$, so any function of $y$ needs the chain rule:

After differentiating, collect the $\frac{dy}{dx}$ terms on one side and factor to solve for the gradient. The result is usually an expression in both $x$ and $y$, which is fine: you supply a point on the curve to get a numeric gradient.

The second derivative and concavity

The second derivative $\frac{d^2y}{dx^2}$ is the rate of change of the gradient. It tells you how the curve bends:

• $\frac{d^2y}{dx^2} > 0$: the gradient is increasing, the curve is concave up (holds water).
• $\frac{d^2y}{dx^2} < 0$: the gradient is decreasing, the curve is concave down.

The second derivative also classifies stationary points: at a stationary point where $\frac{d^2y}{dx^2} > 0$ there is a local minimum, and where $\frac{d^2y}{dx^2} < 0$ a local maximum. If $\frac{d^2y}{dx^2} = 0$ the test is inconclusive and you check the sign of the gradient on each side.

Points of inflection

A point of inflection is where concavity changes from up to down or vice versa. A necessary condition is $\frac{d^2y}{dx^2} = 0$, but that alone is not sufficient: the second derivative must actually change sign there. A point where $\frac{d^2y}{dx^2} = 0$ without a sign change is not an inflection. A stationary point of inflection has both $\frac{dy}{dx} = 0$ and a change of concavity.

Examples in context

Example 1. For $x^2 + xy + y^2 = 7$, differentiating gives $2x + y + x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0$, so $\frac{dy}{dx} = -\frac{2x + y}{x + 2y}$.

Example 2. For $y = x^3$, $\frac{d^2y}{dx^2} = 6x$, which changes sign at $x = 0$, giving a stationary point of inflection at the origin.

Try this

Q1. Find $\frac{dy}{dx}$ for $x^2 - y^2 = 9$ by implicit differentiation. [2 marks]

• Cue. $2x - 2y\frac{dy}{dx} = 0$, so $\frac{dy}{dx} = \frac{x}{y}$.

Q2. State the concavity of $y = e^x$ everywhere. [1 mark]

• Cue. $\frac{d^2y}{dx^2} = e^x > 0$, so concave up everywhere.

Q3. Find any points of inflection of $y = x^3 - 3x$. [3 marks]

• Cue. $\frac{d^2y}{dx^2} = 6x = 0$ at $x = 0$, sign changes, so inflection at $(0, 0)$.