Revise Differentiation for Cambridge 9231 (2.3)

An implicit equation is one that connects $x$ and $y$ without expressing $y$ directly in terms of $x$. A classic example is the equation of a circle, $x² + y² = r²$. While you could solve for $y$ to get $y = ±√(r² - x²)$, this is awkward and creates two separate functions. Implicit differentiation provides a more elegant way to find dy/dx directly from the original equation.

Some curves are best described by expressing both $x$ and $y$ coordinates as functions of a third variable, or parameter, often denoted by $t$. For example, $x = f(t)$ and $y = g(t)$. Think of $t$ as time, and the equations as describing the position of a particle at that time. To find the gradient dy/dx of the particle's path, we can use the chain rule in a specific way.

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$

This formula is derived from the chain rule, $dy/dt = dy/dx * dx/dt$, by rearranging it to make dy/dx the subject. It is valid provided dx/dt ≠ 0.

Finding the second derivative, d²y/dx², is crucial for determining the concavity of a curve and the nature of its stationary points. For parametric equations, this is a multi-step process that is a very common source of error. You cannot simply differentiate the numerator and denominator of dy/dx with respect to $t$.

The correct procedure is to differentiate the expression for dy/dx with respect to $x$, applying the chain rule again:

$\frac{d^2y}{dx^2} = \frac{d}{dx} \left( \frac{dy}{dx} \right) = \frac{d/dt(dy/dx)}{dx/dt}$