# 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.

### Example 2

Find the derivative of

y= (x^{3}− 6x)(2 − 4x^{3})

**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.

Continues below ⇩

## 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)`

### Example 4

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

### 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)`

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