Say we want to differentiate a function of the form $y=\frac{x}{5x-9}$y=x5x−9.
We can see it's in the form of a quotient where we have functions of $x$x in both the numerator and the denominator. While we could apply the product rule by treating $y$y as the product of the functions $u=x$u=x and $v=\left(5x-9\right)^{-1}$v=(5x−9)−1, we will see that the quotient rule in many cases is more efficient.
The quotient rule can be proved from first principles or considered a special case of the product rule and proved using the product and chain rule as follows:
Let $y=\frac{u}{v}$y=uv, where $u$u and $v$v are functions of $x$x.
Firstly, the function $y$y can be rewritten as a product: $y=u\times v^{-1}$y=u×v−1. Applying the product rule we obtain:
| $y'$y′ | $=$= | $u[v^{-1}]'+[v^{-1}]u'$u[v−1]′+[v−1]u′ |
|
| $=$= | $-uv^{-2}v'+v^{-1}u'$−uv−2v′+v−1u′ |
Using the chain rule $[v^{-1}]'=-1v^{-2}\times v'$[v−1]′=−1v−2×v′ |
|
| $=$= | $\frac{-uv'}{v^2}+\frac{u'}{v}$−uv′v2+u′v |
Rewrite as fractions with positive indices |
|
| $=$= | $\frac{-uv'}{v^2}+\frac{vu'}{v^2}$−uv′v2+vu′v2 |
Multiply the second fraction by $\frac{v}{v}$vv to obtain the same denominator |
|
| $=$= | $\frac{vu'-uv'}{v^2}$vu′−uv′v2 |
Combine fractions |
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:
$y'=\frac{vu'-uv'}{v^2}$y′=vu′−uv′v2
Alternatively,
$\frac{dy}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$dydx=vdudx−udvdxv2
| Let | $u=x$u=x | then | $u'=1$u′=1 |
| and | $v=5x-9$v=5x−9 | then | $v'=5$v′=5 |
| $y'$y′ | $=$= | $\frac{vu'-uv'}{v^2}$vu′−uv′v2 |
| $=$= | $\frac{\left(5x-9\right)1-\left(x\right)5}{\left(5x-9\right)^2}$(5x−9)1−(x)5(5x−9)2 | |
| $=$= | $\frac{-9}{\left(5x-9\right)^2}$−9(5x−9)2 |
Could we avoid using the quotient rule and use the product rule instead? Let's compare the methods. We can rewrite the function $y$y to be $y=x\left(5x-9\right)^{-1}$y=x(5x−9)−1.
Using the product rule, $y'=uv'+vu'$y′=uv′+vu′:
| Let | $u=x$u=x | then | $u'=1$u′=1 |
| and | $v=\left(5x-9\right)^{-1}$v=(5x−9)−1 | then | $v'=\dots$v′=… using the chain rule below |
The chain rule tells us to find the derivative of the outside expression and multiply by the derivative of the inside.
Hence,
| $v'$v′ | $=$= | $-1\left(5x-9\right)^{-2}\times5$−1(5x−9)−2×5 |
| $=$= | $-5\left(5x-9\right)^{-2}$−5(5x−9)−2 |
So now we can substitute $u'$u′ and $v'$v′ into the product rule:
| y' | $=$= | $uv'+vu'$uv′+vu′ |
|
| $=$= | $x\left(-5\left(5x-9\right)^{-2}\right)+\left(5x-9\right)^{-1}$x(−5(5x−9)−2)+(5x−9)−1 |
|
|
| $=$= | $-5x\left(5x-9\right)^{-2}+\left(5x-9\right)^{-1}$−5x(5x−9)−2+(5x−9)−1 |
See note below. |
|
| $=$= | $-\frac{5x}{\left(5x-9\right)^2}+\frac{1}{5x-9}$−5x(5x−9)2+15x−9 |
Rewrite as fractions with positive indices. |
|
| $=$= | $-\frac{5x}{\left(5x-9\right)^2}+\frac{\left(5x-9\right)}{\left(5x-9\right)^2}$−5x(5x−9)2+(5x−9)(5x−9)2 |
Rewrite fractions to have a common denominator |
|
| $=$= | $\frac{\left(5x-9\right)-5x}{\left(5x-9\right)^2}$(5x−9)−5x(5x−9)2 |
Combine fractions |
|
| $=$= | $\frac{-9}{\left(5x-9\right)^2}$−9(5x−9)2 |
Simplify |
Note: While we could stop at that stage, it would generally be preferable to write as a single fraction with positive indices.
This process was less efficient and had many steps in the tidying up phase. So while we can apply the product rule, the quotient rule is often more efficient, leaving the answer in a more compact form and less room for error when making simplifications.
Here are a few more examples to consider.
Worked examples
Example 1
Differentiate $y=\frac{3x^2+4}{x^3}$y=3x2+4x3
Think: We consider the numerator and the denominator and make the appropriate designations for $u$u and $v$v in order to use the quotient rule.
Do:
| Let | $u=3x^2+4$u=3x2+4 | then | $u'=6x$u′=6x |
| and | $v=x^3$v=x3 | then | $v'=3x^2$v′=3x2 |
| $y'$y′ | $=$= | $\frac{vu'-uv'}{v^2}$vu′−uv′v2 |
| $=$= | $\frac{x^3\times6x-\left(3x^2+4\right)\left(3x^2\right)}{\left(x^3\right)^2}$x3×6x−(3x2+4)(3x2)(x3)2 | |
| $=$= | $\frac{6x^4-\left(9x^4+12x^2\right)}{x^6}$6x4−(9x4+12x2)x6 | |
| $=$= | $\frac{6x^4-9x^4-12x^2}{x^6}$6x4−9x4−12x2x6 | |
| $=$= | $\frac{-3x^4-12x^2}{x^6}$−3x4−12x2x6 | |
| $=$= | $\frac{-3x^2-12}{x^4}$−3x2−12x4 |
Reflect: Before differentiating we could have rearranged the fraction to be
| $y$y | $=$= | $\frac{3x^2+4}{x^3}$3x2+4x3 |
| $=$= | $\frac{3x^2}{x^3}+\frac{4}{x^3}$3x2x3+4x3 | |
| $=$= | $\frac{3}{x}+\frac{4}{x^3}$3x+4x3 | |
| $=$= | $3x^{-1}+4x^{-3}$3x−1+4x−3 |
Now we can differentiate these two terms using our previous knowledge. This strategy only works because the denominator is just a single term of the form $x^n$xn, so it results in easy to differentiate $x$x terms. The derivative will be:
| $y$y | $=$= | $3x^{-1}+4x^{-3}$3x−1+4x−3 |
This is the simplified expression for y |
| $y'$y′ | $=$= | $-3x^{-2}+\left(-3\right)\times4x^{-4}$−3x−2+(−3)×4x−4 |
Differentiating |
| $=$= | $-3x^{-2}-12x^{-4}$−3x−2−12x−4 |
This is a final answer, but let's make sure it is the same as our original answer. |
|
| $=$= | $\frac{-3}{x^2}-\frac{12}{x^4}$−3x2−12x4 |
Rewrite negative indices as fractions. |
|
| $=$= | $\frac{-3x^2-12}{x^4}$−3x2−12x4 |
Put both fractions over common denominator to get the same answer as before. |
Here, either method can be used but to be able to use the second method the denominator must be a single term involving a power of $x$x.
Example 2
Find the equation of the tangent to the curve $y=\frac{x^2-2}{x+2}$y=x2−2x+2 at the point where $x=2$x=2.
Think: Find the point of contact and the gradient at $x=2$x=2. Then use these to find the equation of the tangent. To find the gradient of the tangent consider the numerator and the denominator and make the appropriate designations for $u$u and $v$v in order to use the quotient rule and find the derivative, then evaluate at $x=2$x=2.
Do:
Point of contact: When $x=2$x=2:
| $y$y | $=$= | $\frac{x^2-2}{x+2}$x2−2x+2 |
| $=$= | $\frac{\left(2\right)^2-2}{2+2}$(2)2−22+2 | |
| $=$= | $\frac{1}{2}$12 |
Hence, the point of contact is $\left(2,\frac{1}{2}\right)$(2,12).
Gradient:
| Let | $u=x^2-2$u=x2−2 | then | $u'=2x$u′=2x |
| and | $v=x+2$v=x+2 | then | $v'=1$v′=1 |
| $y'$y′ | $=$= | $\frac{vu'-uv'}{v^2}$vu′−uv′v2 |
| $=$= | $\frac{\left(x+2\right)2x-\left(x^2-2\right)1}{\left(x+2\right)^2}$(x+2)2x−(x2−2)1(x+2)2 | |
| $=$= | $\frac{2x^2+4x-x^2+2}{\left(x+2\right)^2}$2x2+4x−x2+2(x+2)2 | |
| $=$= | $\frac{x^2+4x+2}{\left(x+2\right)^2}$x2+4x+2(x+2)2 |
When $x=2$x=2:
| $y'$y′ | $=$= | $\frac{\left(2\right)^2+4\left(2\right)+2}{\left(2+2\right)^2}$(2)2+4(2)+2(2+2)2 |
| $=$= | $\frac{16}{16}$1616 | |
| $=$= | $1$1 |
Hence, the gradient of the tangent to the curve at $x=2$x=2 is $m=1$m=1.
Equation of the tangent: Using the point of contact $\left(x_1,y_1\right)=\left(2,\frac{1}{2}\right)$(x1,y1)=(2,12) and gradient $m=1$m=1.
| $y-y_1$y−y1 | $=$= | $m\left(x-x_1\right)$m(x−x1) |
| $y-\frac{1}{2}$y−12 | $=$= | $1\left(x-2\right)$1(x−2) |
| $y$y | $=$= | $x-2+\frac{1}{2}$x−2+12 |
| $y$y | $=$= | $x-\frac{3}{2}$x−32 |
Practice questions
Question 1
Consider the function $y=\frac{3}{x}$y=3x.
By first re-writing with a negative index, find $\frac{dy}{dx}$dydx.
$y=3x^{\editable{}}$y=3x
$\frac{dy}{dx}=\editable{}x^{\editable{}}$dydx=x
Use the quotient rule to differentiate $y=\frac{3}{x}$y=3x.
$\frac{dy}{dx}=\frac{x\times\left(\editable{}\right)-3\times\left(\editable{}\right)}{x^{\editable{}}}$dydx=x×()−3×()x
$\frac{dy}{dx}=\editable{}$dydx=
In which two quadrants of the number plane does the hyperbola $y=\frac{3}{x}$y=3x exist?
$I$I
A$II$II
B$III$III
C$IV$IV
DFor what value of $x$x is the gradient of $y$y undefined?
Question 2
Suppose we want to differentiate $y=\frac{2x+3}{3x-2}$y=2x+33x−2 using the quotient rule.
Using the substitution $u=2x+3$u=2x+3, find $u'$u′.
Using the substitution $v=3x-2$v=3x−2, find $v'$v′.
Hence find $y'$y′.
Is it possible for the derivative of this function to be zero?
No
AYes
B
question 3
Find the value of $f'\left(0\right)$f′(0) if $f\left(x\right)=\frac{x}{\sqrt{16-x^2}}$f(x)=x√16−x2.
You may use the substitutions $u=x$u=x and $v=\sqrt{16-x^2}$v=√16−x2 in your working.