How Do You Differentiate an Implicit Function?

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This is implicit differentiation, covered in IB Mathematics: Analysis and Approaches HL under AHL 5.14: Implicit functions, related rates and optimisation. It is needed when $x$ and $y$ are related by an equation that is inconvenient or impossible to rewrite explicitly as $y=f(x)$.

Why Implicit Differentiation Works

The key mechanism is that $y$ depends on $x$. Therefore, differentiating $y^n$ requires the chain rule:

The factor $\frac{dy}{dx}$ records the rate at which $y$ changes as $x$ changes.

Worked Example

For the circle $x^2+y^2=25$, differentiate every term with respect to $x$:

Then rearrange the differentiated equation:

Thus, the slope at any point on the circle is given by $-\frac{x}{y}$.

When the Product Rule Is Needed

If an equation includes a product containing $y$, use the product rule as well. For

both factors in (xy) can vary with $x$, so its derivative contains two terms. Differentiating gives

Collecting all terms containing $\frac{dy}{dx}$ on one side and solving gives

Gradient at a Point

To find the gradient at $(3,4)$ on the circle, substitute both coordinates into the derivative: $\frac{dy}{dx}=-\frac{3}{4}$. Substituting only $x$ is insufficient because the derivative remains a function of both coordinates.

Common Misconception

The common misconception is to treat $y$ as a constant, giving $\frac{d}{dx}(y^2)=0$. This is wrong because $y$ changes along the curve as $x$ changes. Differentiate each term first, then collect every term containing $\frac{dy}{dx}$.

Exam Technique

In an IB AA HL Paper 1 or Paper 2 response, show the differentiated equation before rearranging. For the command term “find,” relevant stages of working are required; a bare final derivative may not receive full credit. Check that every differentiated $y$-term includes $\frac{dy}{dx}$.