# 6. Derivatives of Products and Quotients

by M. Bourne

## PRODUCT RULE

If *u* and *v* are two functions of *x*, then the derivative of the product *uv* is given by...

### Don't miss...

Later in this section:

Quotient Rule

`(d(uv))/(dx)=u(dv)/(dx)+v(du)/dx`

In words, this can be remembered as:

"The derivative of a product of two functions is the first times the derivative of the second, plus the second times the derivative of the first."

Where does this formula come from? Like all the differentiation formulas we meet, it is based on derivative from first principles.

### Example 1

If we have a product like

y= (2x^{2}+ 6x)(2x^{3}+ 5x^{2})

we can find the derivative without multiplying out the expression on the right.

Answer

We use the substitutions *u* = 2*x*^{2} + 6*x * and
*v *= 2*x*^{3} +
5*x*^{2}.

We can then use the PRODUCT RULE:

`(d(uv))/(dx)=u(dv)/(dx)+v(du)/(dx`

We first find: `(dv)/(dx)=6x^2+10x` and `(du)/(dx)=4x+6`

Then we can write:

`(d(uv))/(dx)=u(dv)/(dx)+v(du)/(dx)`

`=(2x^2+6x)(6x^2+10x)+(2x^3+5x^2)(4x+6)`

`=20x^4+88x^3+90x^2`

Easy to understand math videos:

MathTutorDVD.com

### Example 2

Find the derivative of

y= (x^{3}− 6x)(2 − 4x^{3})

Answer

We recognise this is a product `y=uv`, where

`u=(x^3-6x)` and `v=(2-4x^3)`

So we have:

`(d(uv))/(dx)=u(dv)/(dx)+v(du)/(dx)`

`=(x^3-6x)(-12x^2)+` `(2-4x^3)(3x^2-6)`

`=-12x^5+72x^3+6x^2-` `12-` `12x^5+` `24x^3`

`=-24x^5+96x^3+6x^2-12`

Please support IntMath!

### Why not just multiply them out?

In Example 1 and Example 2, it appears that it would be easier to just expand out the brackets first, then differentiate the result.

However, later we'll need to differentiate functions such as `y = sqrt(x^2-3x)(sin 4x^2)` (in the chapter Differentiation of Transcendental Functions.) It is not possible to multiply this expression term-by-term, so we need a method to differentiate products of such functions.

### Note

We can write the product rule in many different ways:

`(d(uv))/(dx)=uv’+vu’`

OR

`(d(fg))/(dx)` `=f(x)d/(dx)g(x)+g(x)d/(dx)f(x)`… etc.

## QUOTIENT RULE

(A **quotient** is just a fraction.)

If *u* and *v* are two functions of *x*, then the derivative of the quotient `u/v` is given by...

`d/(dx)(u/v)=(v(du)/(dx)-u(dv)/(dx))/(v^2`

In words, this can be remembered as:

"The derivative of a quotient equals bottom times derivative of top minus top times derivative of the bottom, divided by bottom squared."

### Example 3

We wish to find the derivative of the expression:

`y=(2x^3)/(4-x)`

Answer

We recognise that it is in the form: `y=u/v`**.**

We can use the substitutions:

`u = 2x^3` and `v = 4 − x`

Using the **quotient rule,** we first need to find:

`(du)/(dx)=6x^2`

and

`(dv)/(dx)=-1`

Then

`(d(u/v))/(dx)=(v(du)/(dx)-u(dv)/(dx))/v^2`

`=((4-x)(6x^2)-(2x^3)(-1))/((4-x)^2)`

`=(24x^2-6x^3+2x^3)/((4-x)^2)`

`=(24x^2-4x^3)/((4-x)^2)`

### Example 4

Find `(dy)/(dx)` if `y=(4x^2)/(x^3+3)`

Answer

We can use the substitutions:

u= 4x^{2}andv=x^{3}+ 3

Using the **quotient rule,** we first need:

`(du)/(dx)=8x` and `(dv)/(dx)=3x^2`

Then

`(d(u/v))/(dx)=(v(du)/(dx)-u(dv)/(dx))/v^2`

`=((x^3+3)(8x)-(4x^2)(3x^2))/((x^3+3)^2)`

`=(8x^4+24x-12x^4)/((x^3+3)^2)`

`=(-4x^4+24x)/((x^3+3)^2)`

### Interactives

Go to the differentiation applet to explore Examples 3 and 4 and see what we've found.

## Other ways of Writing Quotient Rule

You can also write **quotient rule** as:

`d/(dx)(f/g)=(g\ (df)/(dx)-f\ (dg)/(dx))/(g^2`

OR

`d/(dx)(u/v)=(vu'-uv')/(v^2)`

### Search IntMath, blog and Forum

### Online Calculus Solver

This calculus solver can solve a wide range of math problems.

Go to: Online algebra solver

### Calculus Lessons on DVD

Math videos by MathTutorDVD.com

Easy to understand calculus lessons on DVD. See samples before you commit.

More info: Calculus videos

### The IntMath Newsletter

Sign up for the free **IntMath Newsletter**. Get math study tips, information, news and updates each fortnight. Join thousands of satisfied students, teachers and parents!