Quotient Rule

 

First going back to what we know

Say we want to differentiate a function of the form $y=\frac{x}{5x-9}$y=x5x−9​. 

If we rewrite the function $y$y to be $y=x(5x-9)^{-1}$y=x(5x−9)−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′=u′v+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(5x−9)−2×5=−5(5x−9)−2

 

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

$y'=u'v+uv'$y′=u′v+uv′

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

$y'=(5x-9)^{-1}-5x(5x-9)^{-2}$y′=(5x−9)−1−5x(5x−9)−2

$y'=\frac{1}{5x-9}-\frac{5x}{(5x-9)^2}$y′=15x−9​−5x(5x−9)2​

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

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

$y'=\frac{-9}{(5x-9)^2}$y′=−9(5x−9)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=x5x−9​, we can see it's in the form of a quotient.

The differentiation rule we can use is the quotient rule.

 

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

$y=\frac{x}{5x-9}$y=x5x−9​

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′=u′v−uv′v2​

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

$y'=\frac{-9}{(5x-9)^2}$y′=−9(5x−9)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′=u′v−uv′v2​

$y'=\frac{6x\times x^3-(3x^2+4)(3x^2)}{(x^3)^2}$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′=6x4−9x4−12x2x6​

$y'=\frac{(-3x^4-12x^2)}{x^6}$y′=(−3x4−12x2)x6​

$y'=\frac{(-3x^2-12)}{x^4}$y′=(−3x2−12)x4​

 

More Worked Examples

QUESTION 1

QUESTION 2

QUESTION 3