Calculus How To

Quotient Rule: Definition, Examples

The Quotient Rule

The& quotient rule is used to differentiate functions that are being divided. If you have a function g(x) (top function) divided by h(x) (bottom function) then the quotient rule is:

Formal definition for the quotient rule

Formal definition for the quotient rule.


It looks ugly, but it’s nothing more complicated than following a few steps (which are exactly the same for each quotient). You might also notice that the numerator in the quotient rule is the same as the product rule with one slight difference—the addition sign has been replaced with the subtraction sign.

Watch the video or read on below:

Quotient Rule: Examples

Example Problem #1: Differentiate the following function:
y = 2 / (x + 1)
Solution:
Note: I’m using D as shorthand for derivative here instead of writing g'(x) or f'(x):

  1. Identify g(x) and h(x). The top function (2) is g(x) and the bottom function (x + 1) is f(x).
  2. Plug your functions (from Step 1) into the formula:
    y’ = D{2} (x + 1) – D {x + 1} (2) / (x + 1)2
  3. Work out your derivatives. For example, the derivative of 2 is 0. y’ = (0)(x + 1) – (1)(2) / (x + 1)2
  4. Simplify: y’ = -2 (x + 1)2

When working with the quotient rule, always start with the bottom function, ending with the bottom function squared.

More examples for the Quotient Rule:

  1. How to Differentiate (2x + 1) / (x – 3)
  2. How to Differentiate tan(x)
  3. How to Differentiate 2x / 2xx

How to Differentiate (2x + 1) / (x – 3)

Step 1: Name the top term (the denominator) f(x) and the bottom term (the numerator) g(x). This gives you two new functions:

  • f(x) = 2x + 1
  • g(x) = x – 3

Step 2: Place your functions f(x) and g(x) into the quotient rule. I’ll use d/dx here to indicate a derivative.
f'(x) = (x – 3) d/dx [2x + 1] – (2x + 1) d/dx[x – 3] / [x-3]2

Step 3:Differentiate the indicated functions in Step 2. In this example, those functions are [2x + 1] and [x + 3].
f'(x) = (x – 3)(2)-(2x + 1)(1) / (x – 3)2

Step 4:Use algebra to simplify where possible. The solution is 7/(x – 3)2.

How to Differentiate tan(x)

The quotient rule can be used to differentiate tan(x), because of a basic quotient identity, taken from trigonometry: tan(x) = sin(x) / cos(x).

Step 1: Name the top term f(x) and the bottom term g(x). Using our quotient trigonometric identity tan(x) = sinx(x) / cos(s), then:

  • f(x) = sin(x)
  • g(x) = cos(x)

Step 2: Place your functions f(x) and g(x) into the quotient rule. The term d/dx here indicates a derivative.

f'(x) = cos(x) d/dx[sin(x)] – sin(x) d/dx[cos x]/[cos]2

Step 3:Differentiate the indicated functions from Step 2. In this example, those functions are [sinx(x)] and [cos x].
f'(x)= cos2(x) + sin2(x) / cos2x.

Step 4:Use algebra to simplify where possible. The solution is 1/cos2(x), which is equivalent in trigonometry to sec2(x).

How to Differentiate 2x/2xx

In this example problem, you’ll need to know the algebraic rule that states:
axax = ax + x = a2x and axbx = (ab)x.

Step 1: Name the top term f(x) and the bottom term g(x).

  • f(x) = 2x
  • g(x) = 2x – 3x.

Step 2: Place the functions f(x) and g(x) from Step 1 into the quotient rule. The term d/dx here indicates a derivative.

f'(x) = (2x – 3x) d/dx[2x] – (2x) d/dx[2x – 3x]/(2x
3x)2.

Step 3: Differentiate the indicated functions (d/dx)from Step 2. In this example, those functions are 2x and [2x – 3x]
f'(x)= (2x – 3x) d/dx[2x ln 2] – (2x)(2x2x ln 2 – 3x ln 3).

Step 4: Use algebra to simplify where possible (remembering the rules from the intro).
f'(x) = 22x ln 2 – 6x ln 2 – (22x ln 2 – 6x ln 3) / (2x – 3x)2
By simplification, this becomes:
f'(x) = 6x(ln 3 – ln 2) / (2x-3x)2

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