Quotient Rule for finding Derivatives w/ Step-by-Step Examples

The Quotient Rule helps us to find the derivative of a function that is the quotient of two other functions. It is used when one function is divided by another. This allows us to differentiate these complex functions by considering the individual derivatives of the two functions.

Quotient Rule Formula and Formal Definition

The Quotient Rule formula is as follows:

\frac{d}{d x} (\frac{f (x)}{g (x)}) = \frac{g (x) \cdot f^{'} (x) - f (x) \cdot g^{'} (x)}{[g (x)]^{2}}

For two differentiable functions $f (x)$ and $g (x)$ , where $g (x) \neq 0$ , the derivative of their quotient is given by:

\frac{d}{d x} (\frac{f (x)}{g (x)}) = \frac{g (x) \cdot f^{'} (x) - f (x) \cdot g^{'} (x)}{[g (x)]^{2}}

where $f^{'} (x)$ is the derivative of $f (x)$ and $g^{'} (x)$ is the derivative of $g (x)$ .

Steps to Apply the Quotient Rule

Identify the numerator function, $f (x)$ , and the denominator function, $g (x)$ .
Find the derivative of the numerator function, $f^{'} (x)$ .
Find the derivative of the denominator function, $g^{'} (x)$ .
Apply the Quotient Rule formula:
$\frac{d}{d x} (\frac{f (x)}{g (x)}) = \frac{g (x) \cdot f^{'} (x) - f (x) \cdot g^{'} (x)}{[g (x)]^{2}}$
Simplify the resulting expression, if necessary.

Remembering the Quotient Rule

To easily remember the Quotient Rule, you can use the mnemonic “Hi-Dee-Ho”:

“Hi” represents the numerator function, $f (x)$ , because it’s “high up”.
“Ho” represents the denominator function, $g (x)$ , because it’s “low”.
“Dee” represents the derivative of the function.

The Quotient Rule can be remembered as:

\frac{Hi}{Ho} = \frac{Ho-Dee-Hi - Hi-Dee-Ho}{Ho-Ho}

which translates to:

\frac{g (x) \cdot f^{'} (x) - f (x) \cdot g^{'} (x)}{[g (x)]^{2}}

Examples

Example 1: Derivative of a Fraction

Let’s use the Quotient Rule to differentiate the function $\frac{x^{2} + 1}{x - 1}$ .

Identify $f (x) = x^{2} + 1$ and $g (x) = x - 1$ .
Find $f^{'} (x) = 2 x$ and $g^{'} (x) = 1$ .
Apply the Quotient Rule:
$\frac{d}{d x} (\frac{x^{2} + 1}{x - 1}) = \frac{(x - 1) \cdot 2 x - (x^{2} + 1) \cdot 1}{(x - 1)^{2}}$
Simplify:
$\frac{d}{d x} (\frac{x^{2} + 1}{x - 1}) = \frac{2 x^{2} - 2 x - x^{2} - 1}{(x - 1)^{2}} = \frac{x^{2} - 2 x - 1}{(x - 1)^{2}}$

Example 2: Common Mistake

It’s important to note that the derivative of a quotient is not (!) equal to the quotient of the derivatives. For example:

\frac{d}{d x} (\frac{x^{2}}{x}) \neq \frac{\frac{d}{d x} (x^{2})}{\frac{d}{d x} (x)} = \frac{2 x}{1} = 2 x

Instead, using the Quotient Rule:

\frac{d}{d x} (\frac{x^{2}}{x}) = \frac{x \cdot 2 x - x^{2} \cdot 1}{x^{2}} = \frac{2 x^{2} - x^{2}}{x^{2}} = \frac{x^{2}}{x^{2}} = 1

Example 3: Instantaneous Rate of Change

Find the instantaneous rate of change of $f (x) = \frac{x^{2} + 1}{x + 1}$ at $x = 1$ .

Find $f^{'} (x)$ using the Quotient Rule:
$f^{'} (x) = \frac{(x + 1) \cdot 2 x - (x^{2} + 1) \cdot 1}{(x + 1)^{2}} = \frac{2 x^{2} + 2 x - x^{2} - 1}{(x + 1)^{2}} = \frac{x^{2} + 2 x - 1}{(x + 1)^{2}}$
Evaluate $f^{'} (1)$ :
$f^{'} (1) = \frac{1^{2} + 2 (1) - 1}{(1 + 1)^{2}} = \frac{1 + 2 - 1}{4} = \frac{2}{4} = \frac{1}{2}$

Thus, the instantaneous rate of change of $f (x)$ at $x = 1$ is $\frac{1}{2}$ .

Relationship to the Product Rule

The Quotient Rule is closely related to the Product Rule, which is used to differentiate the product of two functions. The main difference between the two rules is that the Quotient Rule has a minus sign in the numerator, while the Product Rule has a plus sign.

Product Rule:

\frac{d}{d x} (f (x) \cdot g (x)) = f (x) \cdot g^{'} (x) + g (x) \cdot f^{'} (x)

Quotient Rule:

\frac{d}{d x} (\frac{f (x)}{g (x)}) = \frac{g (x) \cdot f^{'} (x) - f (x) \cdot g^{'} (x)}{[g (x)]^{2}}

Even though these functions look similar, it’s really important to use the right rule depending on whether you’re multiplying or dividing them.