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 and the differentiation rule we can use is the quotient rule. The purpose of the quotient rule is to differentiate fractions that have functions of $x$x in both the numerator and the denominator.
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
$\frac{dy}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$dydx=vdudx−udvdxv2
Or, more simply:
$y'=\frac{vu'-uv'}{v^2}$y′=vu′−uv′v2
| Let | $u=x$u=x | and | $v=5x-9$v=5x−9 |
| then | $u'=1$u′=1 | and | $v'=5$v′=5 |
| $y'$y′ | $=$= | $\frac{vu'-uv'}{v^2}$vu′−uv′v2 |
| $y'$y′ | $=$= | $\frac{(5x-9)1-(x)5}{(5x-9)^2}$(5x−9)1−(x)5(5x−9)2 |
| $y'$y′ | $=$= | $\frac{-9}{(5x-9)^2}$−9(5x−9)2 |
As you can see, using the quotient rule is pretty easy, and it generally results in a quite tidy answer. You might be thinking that you don't need to use this rule though, as you could have written the question as a product. Let's compare the methods. We can rewrite the function $y$y to be $y=x(5x-9)^{-1}$y=x(5x−9)−1. Already you can see we have introduced negative powers.
The product rule is: $y'=uv'+vu'$y′=uv′+vu′.
| 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 to 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:
| $y'$y′ | $=$= | $uv'+vu'$uv′+vu′ |
| $y'$y′ | $=$= | $x(-5(5x-9)^{-2})+(5x-9)^{-1}$x(−5(5x−9)−2)+(5x−9)−1 |
| $y'$y′ | $=$= | $-5x(5x-9)^{-2}+(5x-9)^{-1}$−5x(5x−9)−2+(5x−9)−1 |
| $y'$y′ | $=$= | $-\frac{5x}{(5x-9)^2}+\frac{1}{5x-9}$−5x(5x−9)2+15x−9 |
| $y'$y′ | $=$= | $-\frac{5x}{(5x-9)^2}+\frac{(5x-9)}{(5x-9)^2}$−5x(5x−9)2+(5x−9)(5x−9)2 |
| $y'$y′ | $=$= | $\frac{(5x-9)-5x}{(5x-9)^2}$(5x−9)−5x(5x−9)2 |
| $y'$y′ | $=$= | $\frac{-9}{(5x-9)^2}$−9(5x−9)2 |
That was a lot of algebra, mostly in the tidying up steps. Hopefully this might have convinced you that if you are given a question in the form of a fraction leave it as it is, and use the quotient rule, to avoid introducing more unnecessary complexity and a greater margin for error.
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 | and | $v=x^3$v=x3 |
| then | $u'=6x$u′=6x | and | $v'=3x^2$v′=3x2 |
| $y'$y′ | $=$= | $\frac{vu'-uv'}{v^2}$vu′−uv′v2 |
| $y'$y′ | $=$= | $\frac{x^3\times6x-(3x^2+4)(3x^2)}{(x^3)^2}$x3×6x−(3x2+4)(3x2)(x3)2 |
| $y'$y′ | $=$= | $\frac{6x^4-(9x^4+12x^2)}{x^6}$6x4−(9x4+12x2)x6 |
| $y'$y′ | $=$= | $\frac{6x^4-9x^4-12x^2}{x^6}$6x4−9x4−12x2x6 |
| $y'$y′ | $=$= | $\frac{(-3x^4-12x^2)}{x^6}$(−3x4−12x2)x6 |
| $y'$y′ | $=$= | $\frac{-3x^2-12}{x^4}$−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 |
| $y'$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 we want to 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. |
This method would have been much easier and would not have relied on the formula. This will only work in certain cases though.
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: 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 and find the first derivative. We cant simplify the fraction first because the denominator isn'y just one term.
Do:
| Let | $u=x^2-2$u=x2−2 | and | $v=x+2$v=x+2 |
| then | $u'=2x$u′=2x | and | $v'=1$v′=1 |
| $y'$y′ | $=$= | $\frac{vu'-uv'}{v^2}$vu′−uv′v2 |
| $y'$y′ | $=$= | $\frac{(x+2)2x-(x^2-2)1}{(x+2)^2}$(x+2)2x−(x2−2)1(x+2)2 |
| $y'$y′ | $=$= | $\frac{2x^2+4x-x^2+2)}{(x+2)^2}$2x2+4x−x2+2)(x+2)2 |
| $y'$y′ | $=$= | $\frac{x^2+4x+2)}{(x+2)^2}$x2+4x+2)(x+2)2 |
When$x=2$x=2:
$y'=1$y′=1
Equation of the tangent to the curve:
| $y-y_1$y−y1 | $=$= | $m(x-x_1)$m(x−x1) |
| $y-\frac{1}{2}$y−12 | $=$= | $1(x-2)$1(x−2) |
| $y$y | $=$= | $x-2+\frac{1}{2}$x−2+12 |
| $y$y | $=$= | $x-\frac{3}{2}$x−32 |
When applying the quotient rule $y'=\frac{vu'-uv'}{v^2}$y′=vu′−uv′v2 you need to be very careful with the subtraction in the numerator. The expansion will usually involve negative coefficients and this can often lead to errors if you rush. Take it slowly, step by step, showing as much working as possible so that errors (if you make them) are easier to identify.
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 $y=\frac{4x+3}{4x-3}$y=4x+34x−3 using the quotient rule.
Using the substitution $u=4x+3$u=4x+3, find $u'$u′.
Using the substitution $v=4x-3$v=4x−3, find $v'$v′.
Hence find $y'$y′.
Is it possible for the derivative of this function to be zero?
Yes
ANo
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.