The Product and Quotient Rule | A Level Maths Revision Notes

Contents hide

1 Summary

1.1 The Product Rule

1.2 Example #1

1.3 Example #2

1.4 The Quotient Rule

1.5 Example #3

1.5.1 Reference

Summary

The Product Rule Formula:

$\frac { d }{ dx } \left[ f\left( x \right) g\left( x \right) \right] \quad =\quad f^{ ' }\left( x \right) g\left( x \right) \quad +\quad f\left( x \right) g^{ ' }\left( x \right)$

The Quotient Rule Formula:

$\frac { d }{ dx } \left[ \frac { f\left( x \right) }{ g\left( x \right) } \right] \quad =$

$\frac { f^{ ' }\left( x \right) g\left( x \right) \quad -\quad g^{ ' }\left( x \right) f\left( x \right) }{ { \left[ g\left( x \right) \right] }^{ 2 } }$

Where f’(x) and g’(x) are derivatives of f(x) and g(x) respectively.

The Product Rule

When we have to find the derivative of the product of two functions, we apply ”The Product Rule”.

Let’s suppose we have two functions f(x) and g(x).

We are required to find $\frac { d }{ dx } \left[ f\left( x \right) g\left( x \right) \right]$ . Remember, when differentiating a product, each factor is differentiated, but one at a time.

What this basically means is defined by the formula for the product rule.

$\frac { d }{ dx } \left[ f\left( x \right) g\left( x \right) \right] \quad =\quad f^{ ' }\left( x \right) g\left( x \right) \quad +\quad f\left( x \right) g^{ ' }\left( x \right)$

The derivative of the product of two functions is equal to the derivative of the first function multiplied by the second function plus the first function multiplied by the derivative of the second function.

Let’s work out a few examples to understand how this rule is applied.

Example #1

Q. Find the derivative of $2x({ e }^{ x })$

Solution:

We know that:

$f\left( x \right) g\left( x \right) \quad =\quad 2x({ e }^{ x })$

Where:

f(x) = 2x and
$g\left( x \right) \quad =\quad { e }^{ x }$

To Apply the Product Rule Formula:

$\frac { d }{ dx } \left[ f\left( x \right) g\left( x \right) \right] \quad =\quad f^{ ' }\left( x \right) g\left( x \right) \quad +\quad f\left( x \right) g^{ ' }\left( x \right)$

Where f’(x) and g’(x) are derivatives of f(x) and g(x) respectively.

We first find f'(x) and g'(x):

$\frac { d }{ dx } f\left( x \right) \quad =\quad f^{ ' }\left( x \right) \quad =\quad 2$

$\frac { d }{ dx } g\left( x \right) \quad =\quad g^{ ' }\left( x \right) \quad =\quad { e }^{ x }$

$\frac { d }{ dx } { e }^{ x }\quad =\quad { e }^{ x }$

Substitute all the above values into the formula:

$\frac { d }{ dx } \left[ f\left( x \right) g\left( x \right) \right] \quad =\quad 2g\left( x \right) \quad +\quad f\left( x \right) { e }^{ x }$

$=\quad 2\left( { e }^{ x } \right) \quad +\quad (2x){ e }^{ x }$

$=\quad 2{ e }^{ x }\quad +\quad 2x{ e }^{ x }$

Simplify:

Ans: $=\quad 2{ e }^{ x }(1\quad +\quad x)$

Example #2

Q. Find the derivative of $y\quad =\quad x^{ 2 }sin(3x)$

Solution:

We know:

$f\left( x \right) \quad =\quad x^{ 2 }$ and
$g\left( x \right) \quad =\quad sin(3x)$

To Apply the Product Rule Formula:

$\frac { d }{ dx } \left[ f\left( x \right) g\left( x \right) \right] \quad =\quad f^{ ' }\left( x \right) g\left( x \right) \quad +\quad f\left( x \right) g^{ ' }\left( x \right)$

Where f’(x) and g’(x) are derivatives of f(x) and g(x) respectively.

We first find f'(x) and g'(x):

$\frac { d }{ dx } f\left( x \right) \quad =\quad f^{ ' }\left( x \right) \quad =\quad 2x$

$\frac { d }{ dx } g\left( x \right) \quad =\quad g^{ ' }\left( x \right) \quad =\quad 3cos(3x)$

Substitute all the above values into the formula:

$\frac { d }{ dx } \left[ f\left( x \right) g\left( x \right) \right] \quad =\quad 2x\quad g\left( x \right) \quad +\quad f\left( x \right) 3cos(3x)$

$=\quad 2x(sin(3x))\quad +\quad (x^{ 2 })3cos(3x)$

Ans: $=\quad 2xsin(3x)\quad +\quad 3x^{ 2 }cos(3x)$

We applied the product rule when we had to find the derivative of the product of two functions.

However what if we have one function divided by another function. In this case, where derivatives are required of functions that come as quotients we use ”The Quotient Rule”.

The Quotient Rule

In this scenario let’s consider a function which is equal to one function divided by another function i.e.h $h\left( x \right) \quad =\quad \frac { f\left( x \right) }{ g\left( x \right) }$

To solve such functions we use the quotient rule which is defined by the formula:

$\frac { d }{ dx } \left[ \frac { f\left( x \right) }{ g\left( x \right) } \right] \quad =$

$\frac { f^{ ' }\left( x \right) g\left( x \right) \quad -\quad g^{ ' }\left( x \right) f\left( x \right) }{ { \left[ g\left( x \right) \right] }^{ 2 } }$

The derivative of the quotient of two functions is equal to the derivative of the function in the numerator multiplied by the function in the denominator minus the function in the numerator multiplied by the derivative of the function in the denominator and then divide this whole expression by the square of the function in the denominator.

Let’s work out an example to understand how this rule is applied.

Example #3

Q. Find the derivative of $h\left( x \right) \quad =\quad \frac { x\quad -\quad 1 }{ 2x\quad +\quad 1 }$

Solution:

We know:

f(x) = x – 1 and

g(x) = 2x+1

To Apply the Quotient Rule Formula:

$\frac { d }{ dx } \left[ \frac { f\left( x \right) }{ g\left( x \right) } \right] \quad =$

$\frac { f^{ ' }\left( x \right) g\left( x \right) \quad -\quad g^{ ' }\left( x \right) f\left( x \right) }{ { \left[ g\left( x \right) \right] }^{ 2 } }$

Where f’(x) and g’(x) are derivatives of f(x) and g(x) respectively.

We first find f'(x) and g'(x):

$\frac { d }{ dx } f\left( x \right) \quad =\quad f^{ ' }\left( x \right) \quad =\quad 1$

$\frac { d }{ dx } g\left( x \right) \quad =\quad g^{ ' }\left( x \right) \quad =\quad 2$

Substitute all the above values into the formula:

$\frac { d }{ dx } \left[ \frac { f\left( x \right) }{ g\left( x \right) } \right] \quad =\quad \frac { (1)(2x+1)\quad -\quad (2)(x-1) }{ { (2x+1) }^{ 2 } }$

Hence:

$h^{ ' }\left( x \right) \quad =\quad \frac { 2x\quad +\quad 1\quad -\quad 2x\quad +\quad 2 }{ { (2x+1) }^{ 2 } }$

Ans: $h^{ ' }\left( x \right) \quad =\quad \frac { 3 }{ { (2x+1) }^{ 2 } }$