NZ Level 8 (NZC) Level 3 (NCEA) [In development]
Quotient Rule
Lesson

## First going back to what we know

Say we want to differentiate a function of the form $y=\frac{x}{5x-9}$y=x5x9

If we rewrite the function $y$y to be $y=x(5x-9)^{-1}$y=x(5x9)1, we can think of it as a product and set up the product rule for differentiation:

The product rule is: $y'=u'v+uv'$y=uv+uv.

 Let $u=x$u=x and $v=\left(5x-9\right)^{-1}$v=(5x−9)−1 then $u'=1$u′=1 and $v'=...$v′=... we need to use the chain rule

The chain rule tells us : Find the derivative of the outside expression and multiply by the derivative of the inside.

Doing this, we get

$v'=-1(5x-9)^{-2}\times5=-5(5x-9)^{-2}$v=1(5x9)2×5=5(5x9)2

So now we can substitute $u'$u and $v'$v into the product rule:

$y'=u'v+uv'$y=uv+uv

$y'=1(5x-9)^{-1}+x(-5(5x-9)^{-2})$y=1(5x9)1+x(5(5x9)2)

$y'=(5x-9)^{-1}-5x(5x-9)^{-2}$y=(5x9)15x(5x9)2

$y'=\frac{1}{5x-9}-\frac{5x}{(5x-9)^2}$y=15x95x(5x9)2

$y'=\frac{(5x-9)}{(5x-9)^2}-\frac{5x}{(5x-9)^2}$y=(5x9)(5x9)25x(5x9)2

$y'=\frac{(5x-9)-5x}{(5x-9)^2}$y=(5x9)5x(5x9)2

$y'=\frac{-9}{(5x-9)^2}$y=9(5x9)2

WOW! That was a lot of algebra. Mostly in the tidying up, the derivative components were fairly simple.

But you know what - there is another way!

If we look back at the function $y=\frac{x}{5x-9}$y=x5x9, we can see it's in the form of a quotient.

The differentiation rule we can use is the quotient rule.

Quotient Rule

If a function is of the form $y=\frac{u}{v}$y=uv, where $u$u and $v$v are functions of $x$x, then

$y'=\frac{u'v-uv'}{v^2}$y=uvuvv2

So for the example above, we could have found the derivative this way.

$y=\frac{x}{5x-9}$y=x5x9

 Let $u=x$u=x and $v=5x-9$v=5x−9 then $u'=1$u′=1 and $v'=5$v′=5

$y'=\frac{u'v-uv'}{v^2}$y=uvuvv2

$y'=\frac{1(5x-9)-(x)5}{(5x-9)^2}$y=1(5x9)(x)5(5x9)2

Already, we have the nice denominator that took us a few lines of algebra to get to before!

$y'=\frac{(5x-9)-5x}{(5x-9)^2}$y=(5x9)5x(5x9)2

$y'=\frac{-9}{(5x-9)^2}$y=9(5x9)2

That seems a whole lot easier!

#### Another Example

Differentiate $y=\frac{3x^2+4}{x^3}$y=3x2+4x3

 Let $u=3x^2+4$u=3x2+4 and $v=x^3$v=x3 then $u'=6x$u′=6x and $v'=3x^2$v′=3x2

$y'=\frac{u'v-uv'}{v^2}$y=uvuvv2

$y'=\frac{6x\times x^3-(3x^2+4)(3x^2)}{(x^3)^2}$y=6x×x3(3x2+4)(3x2)(x3)2

$y'=\frac{6x^4-(9x^4+12x^2)}{x^6}$y=6x4(9x4+12x2)x6

$y'=\frac{6x^4-9x^4-12x^2}{x^6}$y=6x49x412x2x6

$y'=\frac{(-3x^4-12x^2)}{x^6}$y=(3x412x2)x6

$y'=\frac{(-3x^2-12)}{x^4}$y=(3x212)x4

Be careful!

When applying the quotient rule $y'=\frac{u'v-uv'}{v^2}$y=uvuvv2

Be very careful with the subtraction in the numerator - this often creates an expansion involving negative coefficients that often leads to student errors.

Take them slowly, step by step, setting out as much work as possible so that errors (if you make them) are easier to identify.

#### More Worked Examples

##### QUESTION 2

Suppose we want to differentiate $y=\frac{2x+3}{3x-2}$y=2x+33x2 using the quotient rule.

1. Using the substitution $u=2x+3$u=2x+3, find $u'$u.

2. Using the substitution $v=3x-2$v=3x2, find $v'$v.

3. Hence find $y'$y.

4. Is it possible for the derivative of this function to be zero?

No

A

Yes

B

No

A

Yes

B

##### QUESTION 3

Consider the function $y=\frac{5x}{3x-4}$y=5x3x4.

1. Differentiate $y$y.

2. Is it possible for the derivative to be zero?

No

A

Yes

B

No

A

Yes

B
3. Solve for the value of $x$x that will make the derivative undefined.

4. Is the function increasing or decreasing in each of the intervals $\left(-\infty,\frac{4}{3}\right)$(,43) and $\left(\frac{4}{3},\infty\right)$(43,)?

Decreasing

A

Increasing

B

Decreasing

A

Increasing

B

### Outcomes

#### M8-11

Choose and apply a variety of differentiation, integration, and antidifferentiation techniques to functions and relations, using both analytical and numerical methods

#### 91578

Apply differentiation methods in solving problems