7.3.5 Quotient Rule

The quotient rule is a differentiation technique used to find the derivative of a quotient of two functions. It's particularly useful when dealing with functions that are divided by one another, and it complements the product rule and chain rule in calculus.

1. The Quotient Rule Formula:

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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.

2. Understanding the Quotient Rule:

The quotient rule helps to account for how both the numerator and the denominator change. It subtracts the product of the derivative of the numerator and the denominator from the product of the numerator and the derivative of the denominator, and then divides the result by the square of the denominator.

3. Steps to Apply the Quotient Rule:

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

4. Examples of Applying the Quotient Rule:

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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:

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

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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:

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

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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:

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

5. Combining the Quotient Rule with the Chain Rule:

Sometimes, the quotient rule must be used alongside the chain rule, especially when the functions involved are themselves composite functions.

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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:

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

Summary:

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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.

Quotient Rule for Differentiation

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

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Proof:

Start with the definition:

\frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \lim_{h \to 0} \left(\frac{\frac{u(x+h)}{v(x+h)} - \frac{u(x)}{v(x)}}{h}\right)

Combine the fractions:

\lim_{h \to 0} \left[\frac {1}{h} \times \left(\frac{u(x+h)}{v(x+h)} - \frac{u(x)}{v(x)}\right) \right]

Factor and rearrange:

= \lim_{h \to 0} \left( \frac{1}{h} \times \frac{u(x+h)v(x) - u(x)v(x+h)}{v(x+h)v(x)} \right)

Further expand:

= \lim_{h \to 0} \left( \frac{1}{v(x+h)v(x)} \times \frac{u(x+h)v(x) - u(x)v(x+h)}{h} \right)

= \lim_{{h \to 0}} \left( \frac{1}{{v(x+h)v(x)}} \times \frac{u(x+h)v(x) - v(x)u(x) + v(x)u(x) - u(x)v(x+h)}{h} \right)

= \lim_{{h \to 0}} \left( \frac{1}{{v(x+h)v(x)}} \right) \times \frac{v(x)[u(x+h) - u(x)] + u(x)[v(x) - v(x+h)]}{h}

= \lim_{{h \to 0}} \left( \frac{1}{{v(x+h)v(x)}} \right) \times \frac{v(x)[u(x+h) - u(x)] - u(x)[v(x+h) - v(x)]}{h}

= \lim_{{h \to 0}} \left( \frac{1}{{v(x+h)v(x)}} \right) \times \left[ \lim_{{h \to 0}} \left( v(x) \frac{u(x+h) - u(x)}{h} \right) - \lim_{{h \to 0}} \left( u(x) \frac{v(x+h) - v(x)}{h} \right) \right]

Split the limit and apply limit laws:

= \frac{1}{[v(x)]^2} \times \left[ v(x) \lim_{h \to 0} \left (\frac{u(x+h) - u(x)}{h}\right) - u(x) \lim_{h \to 0} \left (\frac{v(x+h) - v(x)}{h}\right ) \right]

\frac{1}{[v(x)]^2} \times \left[ v(x)u'(x) - u(x)v'(x) \right]

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

Thus, the proof for the quotient rule is completed:

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

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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 $v' = 3x^2$

\left (\frac {u}{v}\right )^' = \frac {vu'-uv'}{v^2}

Therefore,

\frac{dy}{dx} = \frac{(x^2 - 3)(2x) - (x^2 + 2)(3x^2)}{(x^2 - 3)^2}

When $x = 2$ :

\frac{dy}{dx} = \frac{(2^2 - 3)(4) - (2^2 + 2)(3 \times 2^2)}{(2^2 - 3)^2} = -2.08

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

y = \frac{2^2 + 2}{2^3 - 3} = 1.2

Therefore, the equation of the tangent is:

y - 1.2 = -2.08(x - 2)

y - 1.2 = -2.08x + 4.16

y = -2.08x + 5.36

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(Jun 2005, Q6)

(a) Find the exact value of the $x$ -coordinate of the stationary point of the curve $y = x \ln x$ .

Let $u = x$ , so $u' = 1$ .

Let $v = \ln x$ , so $v' = \frac{1}{x}$ .

\frac{dy}{dx} = \ln x + 1 = 0

\ln x = -1 \Rightarrow x = e^{-1}

(b) The equation of a curve is $y = \frac{4x + c}{4x - c}$ , where c is a non-zero constant. Show by differentiation that this curve has no stationary points.

Let $u = 4x + c$ , so $u' = 4$ .

Let $v = 4x - c$ , so $v' = 4$ .

\frac{dy}{dx} = \frac{(4x-c)(4) - (4x+c)(4)}{(4x-c)^2} \Rightarrow \frac {16x-4c-16x-4c}{(4x-c)^2}\Rightarrow \frac{-8c}{(4x-c)^2}\Rightarrow 0

Stationary points occur when $c = 0$ , but $c$ cannot be $0$ , so there are no stationary points.

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Quotient Rule Simplified Revision Notes for A-Level OCR Maths Pure

7.3.5 Quotient Rule

1. The Quotient Rule Formula:

2. Understanding the Quotient Rule:

3. Steps to Apply the Quotient Rule:

4. Examples of Applying the Quotient Rule:

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

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

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

5. Combining the Quotient Rule with the Chain Rule:

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

Summary:

Quotient Rule for Differentiation

Proof:

500K+ Students Use These Powerful Tools to Master Quotient Rule For their A-Level Exams.

Other Revision Notes related to Quotient Rule you should explore

Quotient Rule Simplified Revision Notes for A-Level OCR Maths Pure

7.3.5 Quotient Rule

1. The Quotient Rule Formula:

2. Understanding the Quotient Rule:

3. Steps to Apply the Quotient Rule:

4. Examples of Applying the Quotient Rule:

Example 1: Differentiate y=x2+1x2−1y = \frac{x^2 + 1}{x^2 - 1}y=x2−1x2+1​

Example 2: Differentiate y=sin⁡(x)xy = \frac{\sin(x)}{x}y=xsin(x)​

Example 3: Differentiate y=exln⁡(x)y = \frac{e^x}{\ln(x)}y=ln(x)ex​

5. Combining the Quotient Rule with the Chain Rule:

Example: Differentiate y=x2e3xy = \frac{x^2}{e^{3x}}y=e3xx2​

Summary:

Quotient Rule for Differentiation

Proof:

500K+ Students Use These Powerful Tools to Master Quotient Rule For their A-Level Exams.

Other Revision Notes related to Quotient Rule you should explore

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

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

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

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