How Do You Use the Quotient Rule?

Use the quotient rule when differentiating one function divided by another. If $y=\frac{u(x)}{v(x)}$, where $v(x)\neq 0$, then

$$\frac{dy}{dx}=\frac{v(x)u'(x)-u(x)v'(x)}{[v(x)]^2}.$$

The order in the numerator matters: denominator times derivative of numerator, minus numerator times derivative of denominator.

This rule belongs to IB Mathematics: Analysis and Approaches SL 5.6: Differentiation rules. It is required at both SL and HL and can be examined on Paper 1 or Paper 2.

How the Rule Works

1. Identify the numerator as $u(x)$ and the denominator as $v(x)$.
2. Differentiate both separately to obtain (u'(x)) and (v'(x)).
3. Substitute carefully into $\frac{vu'-uv'}{v^2}$.

For example, differentiate

$$f(x)=\frac{x^2+1}{x-2}.$$

Let $u=x^2+1$ and $v=x-2$. Therefore, $u'=2x$ and . Substitution gives

$$f'(x)=\frac{(x-2)(2x)-(x^2+1)(1)}{(x-2)^2}.$$

Expanding the numerator produces

$$f'(x)=\frac{2x^2-4x-x^2-1}{(x-2)^2}=\frac{x^2-4x-1}{(x-2)^2}.$$

The quotient rule is needed because both parts of the quotient vary with $x$. It follows from rewriting $y=uv^{-1}$ and applying the product rule together with the chain rule.

The mechanism can be seen by rewriting the quotient as a product. Since $y=uv^{-1}$, the product rule gives $y'=u'v^{-1}+u(-v^{-2}v')$. Placing both terms over the common denominator gives . This explains both the subtraction sign and the squared denominator.

Before using the rule, check whether algebraic simplification makes differentiation easier. For example, $\frac{x^2}{x}=x$ for $x\neq 0$, so its derivative is simply on that domain. However, do not cancel terms across addition: cannot be reduced by cancelling . In that case, use the quotient rule or rewrite it as .

Common Errors

IB Exam Technique

Write the quotient-rule substitution before simplifying. This demonstrates a valid method and may earn a method mark even if later algebra is incorrect. Keep brackets around each product, check subtraction signs, and simplify only at the end.