6. Derivatives of Products and Quotients
by M. Bourne
PRODUCT RULE
If u and v are two functions of x, then the derivative of the product uv is given by...
Don't miss...
Later in this section:
Quotient Rule
`(d(uv))/(dx)=u(dv)/(dx)+v(du)/dx`
In words, this can be remembered as:
"The derivative of a product of two functions is the first times the derivative of the second, plus the second times the derivative of the first."
Where does this formula come from? Like all the differentiation formulas we meet, it is based on derivative from first principles.
Example 1
If we have a product like
y = (2x2 + 6x)(2x3 + 5x2)
we can find the derivative without multiplying out the expression on the right.
Answer
We use the substitutions u = 2x2 + 6x and v = 2x3 + 5x2.
We can then use the PRODUCT RULE:
`(d(uv))/(dx)=u(dv)/(dx)+v(du)/(dx`
We first find: `(dv)/(dx)=6x^2+10x` and `(du)/(dx)=4x+6`
Then we can write:
`(d(uv))/(dx)=u(dv)/(dx)+v(du)/(dx)`
`=(2x^2+6x)(6x^2+10x)+(2x^3+5x^2)(4x+6)`
`=20x^4+88x^3+90x^2`
Example 2
Find the derivative of
y = (x3 − 6x)(2 − 4x3)
Answer
We recognise this is a product `y=uv`, where
`u=(x^3-6x)` and `v=(2-4x^3)`
So we have:
`(d(uv))/(dx)=u(dv)/(dx)+v(du)/(dx)`
`=(x^3-6x)(-12x^2)+` `(2-4x^3)(3x^2-6)`
`=-12x^5+72x^3+6x^2-` `12-` `12x^5+` `24x^3`
`=-24x^5+96x^3+6x^2-12`
Why not just multiply them out?
In Example 1 and Example 2, it appears that it would be easier to just expand out the brackets first, then differentiate the result.
However, later we'll need to differentiate functions such as `y = sqrt(x^2-3x)(sin 4x^2)` (in the chapter Differentiation of Transcendental Functions.) It is not possible to multiply this expression term-by-term, so we need a method to differentiate products of such functions.
Note
We can write the product rule in many different ways:
`(d(uv))/(dx)=uv’+vu’`
OR
`(d(fg))/(dx)` `=f(x)d/(dx)g(x)+g(x)d/(dx)f(x)`… etc.
QUOTIENT RULE
(A quotient is just a fraction.)
If u and v are two functions of x, then the derivative of the quotient `u/v` is given by...
`d/(dx)(u/v)=(v(du)/(dx)-u(dv)/(dx))/(v^2`
In words, this can be remembered as:
"The derivative of a quotient equals bottom times derivative of top minus top times derivative of the bottom, divided by bottom squared."
Example 3
We wish to find the derivative of the expression:
`y=(2x^3)/(4-x)`
Answer
We recognise that it is in the form: `y=u/v`.
We can use the substitutions:
`u = 2x^3` and `v = 4 − x`
Using the quotient rule, we first need to find:
`(du)/(dx)=6x^2`
and
`(dv)/(dx)=-1`
Then
`(d(u/v))/(dx)=(v(du)/(dx)-u(dv)/(dx))/v^2`
`=((4-x)(6x^2)-(2x^3)(-1))/((4-x)^2)`
`=(24x^2-6x^3+2x^3)/((4-x)^2)`
`=(24x^2-4x^3)/((4-x)^2)`
Example 4
Find `(dy)/(dx)` if `y=(4x^2)/(x^3+3)`
Answer
We can use the substitutions:
u = 4x2 and v = x3 + 3
Using the quotient rule, we first need:
`(du)/(dx)=8x` and `(dv)/(dx)=3x^2`
Then
`(d(u/v))/(dx)=(v(du)/(dx)-u(dv)/(dx))/v^2`
`=((x^3+3)(8x)-(4x^2)(3x^2))/((x^3+3)^2)`
`=(8x^4+24x-12x^4)/((x^3+3)^2)`
`=(-4x^4+24x)/((x^3+3)^2)`
Interactives
Go to the differentiation applet to explore Examples 3 and 4 and see what we've found.
Other ways of Writing Quotient Rule
You can also write quotient rule as:
`d/(dx)(f/g)=(g\ (df)/(dx)-f\ (dg)/(dx))/(g^2`
OR
`d/(dx)(u/v)=(vu'-uv')/(v^2)`
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