Quotient Rule and Rational Functions

Quotient Rule and Rational Functions

A rational function is a quotient of two polynomials. To differentiate it we need the quotient rule — a formula that handles the numerator and denominator simultaneously. On this page we learn the formula, understand where the term order and the minus sign in the numerator come from, and practice evaluating the derivative at a specific point.

Background and Basic Definitions

A rational function is a quotient of two expressions: \(f(x)=\dfrac{u(x)}{v(x)}\), where \(v(x)\neq 0\).

Quotient Rule:

\[ f'(x)=\frac{u'(x)\,v(x)-u(x)\,v'(x)}{\big(v(x)\big)^2} \]

Two things to note: the order in the numerator matters — start with \(u'v\) and then subtract \(uv'\); the sign is minus, so you must not swap the order of the terms. The denominator is the square of the original denominator.

A helpful mnemonic: 'derivative of the top times the bottom, minus the top times the derivative of the bottom, all over the bottom squared'.

Recall the tools: \((x^n)'=n x^{n-1}\), \((c)'=0\), \((ax+b)'=a\). To evaluate the derivative at a point, first find the expression \(f'(x)\), then substitute the \(x\)-value.

Basic Derivatives Table:

FunctionDerivativeFunctionDerivative
\(x^t\)\(t\cdot x^{t-1}\)\(\frac{1}{x}\)\(-\frac{1}{x^2}\)
\(\sqrt{x}\)\(\frac{1}{2\sqrt{x}}\)\(a^x\)\(a^x\cdot \ln a\)
\(\sin x\)\(\cos x\)\(\cos x\)\(-\sin x\)
\(\tan x\)\(\frac{1}{\cos^2 x}\)\(\log_a x\)\(\frac{1}{x\cdot \ln a}\)

where \(t\) is any real number. Differentiation rules: product \([f\cdot g]'=f'g+fg'\); quotient \(\left[\frac{f}{g}\right]'=\frac{f'g-fg'}{[g]^2}\); chain rule \([f(u(x))]'=f'(u)\cdot u'(x)\).

Solution Steps

  1. Step 1 — Label the numerator \(u(x)\) and the denominator \(v(x)\).
  2. Step 2 — Differentiate each separately to get \(u'(x)\) and \(v'(x)\).
  3. Step 3 — Substitute into the formula \(f'=\dfrac{u'v-uv'}{v^2}\), paying careful attention to the order and the minus sign in the numerator.
  4. Step 4 — Expand and simplify the numerator (the denominator is usually left as a square).
  5. Step 5 — If \(f'(a)\) is requested, substitute \(a\) now and compute the numerator and denominator separately.
  6. Step 6 — Verify that the denominator at the point is not zero; if it is, the function is undefined there.

Worked Examples

Example 1: Quotient Rule — Evaluating the Derivative at a Point

Problem: Given the rational function \(f(x)=\dfrac{x^2+3}{x-1}\). Find \(f'(2)\).

Solution:

  1. Let \(u(x)=x^2+3\) and \(v(x)=x-1\).
  2. Differentiate: \(u'(x)=2x\) and \(v'(x)=1\).
  3. Apply the quotient rule: \(f'(x)=\dfrac{2x\,(x-1)-(x^2+3)\cdot 1}{(x-1)^2}\).
  4. Substitute \(x=2\): numerator \(=2\cdot 2\,(2-1)-(2^2+3)=4\cdot 1-7=-3\); denominator \(=(2-1)^2=1\).
  5. Divide: \(f'(2)=\dfrac{-3}{1}=-3\).

Answer: \(f'(2)=-3\).

Example 2: Quotient of Two Linear Expressions

Problem: Given the function \(f(x)=\dfrac{2x+1}{x+2}\). Find \(f'(1)\).

Solution:

  1. Let \(u(x)=2x+1\) and \(v(x)=x+2\).
  2. Differentiate: \(u'(x)=2\) and \(v'(x)=1\).
  3. Substitute: \(f'(x)=\dfrac{2\,(x+2)-(2x+1)\cdot 1}{(x+2)^2}\).
  4. Simplify the numerator: \(2x+4-2x-1=3\), so \(f'(x)=\dfrac{3}{(x+2)^2}\).
  5. Substitute \(x=1\): \(f'(1)=\dfrac{3}{(1+2)^2}=\dfrac{3}{9}=\dfrac{1}{3}\).

Answer: \(f'(1)=\dfrac{1}{3}\).

Example 3: Quadratic Numerator, Evaluating at a Point

Problem: Given the function \(f(x)=\dfrac{x^2-4}{x+1}\). Find \(f'(0)\).

Solution:

  1. Let \(u(x)=x^2-4\) and \(v(x)=x+1\).
  2. Differentiate: \(u'(x)=2x\) and \(v'(x)=1\).
  3. Substitute: \(f'(x)=\dfrac{2x\,(x+1)-(x^2-4)\cdot 1}{(x+1)^2}\).
  4. Substitute \(x=0\): numerator \(=2\cdot 0\,(0+1)-(0-4)=0-(-4)=4\); denominator \(=(0+1)^2=1\).
  5. Divide: \(f'(0)=\dfrac{4}{1}=4\).

Answer: \(f'(0)=4\).

Example 4: Linear Quotient — Checking the Minus Sign

Problem: Given the function \(f(x)=\dfrac{3x-2}{2x+1}\). Find \(f'(0)\).

Solution:

  1. Let \(u(x)=3x-2\) and \(v(x)=2x+1\).
  2. Differentiate: \(u'(x)=3\) and \(v'(x)=2\).
  3. Substitute: \(f'(x)=\dfrac{3\,(2x+1)-(3x-2)\cdot 2}{(2x+1)^2}\).
  4. Simplify the numerator: \(6x+3-6x+4=7\), so \(f'(x)=\dfrac{7}{(2x+1)^2}\).
  5. Substitute \(x=0\): \(f'(0)=\dfrac{7}{(2\cdot 0+1)^2}=\dfrac{7}{1}=7\).

Answer: \(f'(0)=7\).

Example 5: Quadratic Numerator, Non-Cancelling Denominator

Problem: Given the function \(f(x)=\dfrac{x^2+9}{x+3}\). Find \(f'(0)\).

Solution:

  1. Let \(u(x)=x^2+9\) and \(v(x)=x+3\).
  2. Differentiate: \(u'(x)=2x\) and \(v'(x)=1\).
  3. Substitute: \(f'(x)=\dfrac{2x\,(x+3)-(x^2+9)\cdot 1}{(x+3)^2}\).
  4. Substitute \(x=0\): numerator \(=0\,(0+3)-(0+9)=-9\); denominator \(=(0+3)^2=9\).
  5. Divide: \(f'(0)=\dfrac{-9}{9}=-1\).

Answer: \(f'(0)=-1\).

Common Mistakes

✗ Common mistake: Swapping the order in the numerator and writing \(uv'-u'v\) instead of \(u'v-uv'\).

✓ The correct way: The order is fixed: first 'derivative of the numerator times the denominator', then subtract 'numerator times derivative of the denominator'. Swapping the order flips the sign of the result.

✗ Common mistake: Forgetting to square the denominator and leaving \(v(x)\) instead of \(\big(v(x)\big)^2\).

✓ The correct way: The denominator in the quotient rule is always the square of the original denominator. Make sure you have written \((v(x))^2\).

✗ Common mistake: Substituting the value of the point at the very beginning instead of finding \(f'(x)\) first.

✓ The correct way: First find the general derivative expression, then substitute \(x\). Compute the numerator and denominator separately to avoid sign errors.

Practice Tips

  • Tip — Memorize the phrase: 'derivative of the top times the bottom, minus the top times the derivative of the bottom, all over the bottom squared'.
  • Tip — When both numerator and denominator are linear, the numerator of the derivative always simplifies to a constant, and the derivative becomes \(\dfrac{\text{constant}}{(\text{denominator})^2}\).
  • Tip — Before substituting, confirm that the denominator at the point is not zero; if it is, the function is undefined there and no derivative exists.
  • Tip — Keep parentheses around every expression, especially after the minus sign — that is the most common source of sign errors.

Summary and Key Formulas

Quotient Rule:

\[ \left(\frac{u}{v}\right)' = \frac{u'v-uv'}{v^2} \]
  • The numerator order is fixed: \(u'v\) minus \(uv'\).
  • The denominator is the square of the original denominator.
  • To find \(f'(a)\): obtain \(f'(x)\) first, then substitute at the end.
  • Verify that \(v(a)\neq 0\).