Differentiation

Lesson

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

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

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(5`x`−9)−2×5=−5(5`x`−9)−2

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

$y'=u'v+uv'$`y`′=`u`′`v`+`u``v`′

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

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

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

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

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

$y'=\frac{-9}{(5x-9)^2}$`y`′=−9(5`x`−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`=`x`5`x`−9, 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`=`u``v`, where $u$`u` and $v$`v` are functions of $x$`x`, then

$y'=\frac{u'v-uv'}{v^2}$`y`′=`u`′`v`−`u``v`′`v`2

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

$y=\frac{x}{5x-9}$`y`=`x`5`x`−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`−`u``v`′`v`2

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

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

That seems a whole lot easier!

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

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`−`u``v`′`v`2

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

$y'=\frac{6x^4-(9x^4+12x^2)}{x^6}$`y`′=6`x`4−(9`x`4+12`x`2)`x`6

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

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

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

Be careful!

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

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.

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

Using the substitution $u=2x+3$

`u`=2`x`+3, find $u'$`u`′.Using the substitution $v=3x-2$

`v`=3`x`−2, find $v'$`v`′.Hence find $y'$

`y`′.Is it possible for the derivative of this function to be zero?

No

AYes

BNo

AYes

B

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

Differentiate $y$

`y`.Is it possible for the derivative to be zero?

No

AYes

BNo

AYes

BSolve for the value of $x$

`x`that will make the derivative undefined.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

AIncreasing

BDecreasing

AIncreasing

B

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

Apply differentiation methods in solving problems