Quotient Rule (AQA A-Level Mathematics): Revision Notes

If you have two functions $u(x)$ and $v(x)$, and you want to differentiate the quotient $\frac{u(x)}{v(x)}$ with respect to x, the quotient rule states:

$\frac{d}{dx} \left(\frac{u(x)}{v(x)}\right) = \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{[v(x)]^2}$

• $u(x)$ is the numerator.
• $v(x)$ is the denominator.
• $u'(x)$ is the derivative of the numerator.
• $v'(x)$ is the derivative of the denominator.

1. Identify the Functions: Determine the functions $u(x)$ (numerator) and $v(x)$ (denominator).
2. Differentiate Each Function: Find the derivatives $u'(x)$ and $v'(x)$.
3. Substitute into the Quotient Rule Formula: Plug the derivatives into the quotient rule formula.
4. Simplify: Simplify the expression, if possible.

Example 1: Differentiate $y = \frac{x^2 + 1}{x^2 - 1}$

• Step 1: Identify the functions:
• Numerator: $u(x) = x^2 + 1$
• Denominator: $v(x) = x^2 - 1$
• Step 2: Differentiate each function:
• $u'(x) = 2x$
• $v'(x) = 2x$
• Step 3: Apply the quotient rule:

$\frac{dy}{dx} = \frac{(x^2 - 1) \cdot 2x - (x^2 + 1) \cdot 2x}{(x^2 - 1)^2}$

• Step 4: Simplify:

$\frac{dy}{dx} = \frac{2x(x^2 - 1 - x^2 - 1)}{(x^2 - 1)^2} = \frac{2x(-2)}{(x^2 - 1)^2} = \frac{-4x}{(x^2 - 1)^2}$

So, the derivative is:

$\frac{dy}{dx} = \frac{-4x}{(x^2 - 1)^2}$

Example 2: Differentiate $y = \frac{\sin(x)}{x}$

• Step 1: Identify the functions:
• Numerator: $u(x) = \sin(x)$
• Denominator: $v(x) = x$
• Step 2: Differentiate each function:
• $u'(x) = \cos(x)$
• $v'(x) = 1$
• Step 3: Apply the quotient rule:

$\frac{dy}{dx} = \frac{x \cdot \cos(x) - \sin(x) \cdot 1}{x^2}$

• Step 4: Simplify:

$\frac{dy}{dx} = \frac{x\cos(x) - \sin(x)}{x^2}$

Example 3: Differentiate $y = \frac{e^x}{\ln(x)}$

• Step 1: Identify the functions:
• Numerator: $u(x) = e^x$
• Denominator: $v(x) = \ln(x)$
• Step 2: Differentiate each function:
• $u'(x) = e^x$
• $v'(x) = \frac{1}{x}$
• Step 3: Apply the quotient rule:

$\frac{dy}{dx} = \frac{\ln(x) \cdot e^x - e^x \cdot \frac{1}{x}}{[\ln(x)]^2}$

• Step 4: Simplify:

$\frac{dy}{dx} = \frac{e^x \left(\ln(x) - \frac{1}{x}\right)}{[\ln(x)]^2}$

Example: Differentiate $y = \frac{x^2}{e^{3x}}$

• Step 1: Identify the functions:
• Numerator: $u(x) = x^2$
• Denominator: $v(x) = e^{3x}$
• Step 2: Differentiate each function:
• $u'(x) = 2x$
• $v'(x) = 3e^{3x}$ (using the chain rule)
• Step 3: Apply the quotient rule:

$\frac{dy}{dx} = \frac{e^{3x} \cdot 2x - x^2 \cdot 3e^{3x}}{(e^{3x})^2}$

• Step 4: Simplify:

$\frac{dy}{dx} = \frac{e^{3x}(2x - 3x^2)}{e^{6x}} = \frac{2x - 3x^2}{e^{3x}}$

• The quotient rule is essential for differentiating functions that are ratios of two other functions.
• By carefully applying the rule and then simplifying, you can find the derivative of even complex quotient functions.
• Mastering the quotient rule, along with other differentiation techniques, is crucial for solving a wide range of problems in calculus and its applications.

Proof:

Start with the definition:

Combine the fractions:

Factor and rearrange:

Further expand:

Split the limit and apply limit laws:

Thus, the proof for the quotient rule is completed:

Example: Find the equation of the tangent to $y = \dfrac{x^2 + 2}{x^2 - 3}$ when $x = 2$

• Let $u = x^2 + 2$, so $u' = 2x$
• Let $v = x^2 - 3$, so $v' = 3x^2$

Therefore,

When $x = 2$:

Substitute $x = 2$ into the original function to find $y$:

Therefore, the equation of the tangent is:

(Jun 2005, Q6)