Today - The Product RuleToday - The Product RuleDiff functions that are Diff functions that are

productsproductsLikeLike

© Christine Crisp

23 )4( xLast lesson – Chain RuleLast lesson – Chain Rule

)1( 210 xxy

The Product RuleThe product rule gives us a way of

differentiating functions which are multiplied together.Consider (a) )1( 210 xxy

and (b)

102 )1( xxy

However, doing (a) gives us a clue for a new method.

but, we need an easier way of doing (b) since

has 11 terms!10)1( x

(a) can be differentiated by multiplying out the brackets . . .

1012 xxy

The Product Rule

)1( 210 xxy(a)

911 1012 xx 1012 xxy Multiplying

out:

dx

dy

Now suppose we differentiate the 2 functions in without multiplying them out.

)1( 210 xx

Multiplying these 2 answers does NOT give dx

dyBUT

dx

duv

910xdx

du

dx

dvx2

Let and10x 12 xu v

The Product Rule

)1( 210 xxy(a)

911 1012 xx 1012 xxy Multiplying

out:

dx

dy

Now suppose we differentiate the 2 functions in without multiplying them out.

)1( 210 xx

BUT dx

duv

910xdx

du

dx

dvx2

Let and10x 12 xu v

911 1010 xx and dx

dvu

The Product Rule

)1( 210 xxy(a)

911 1012 xx 1012 xxy Multiplying

out:

dx

dy

Now suppose we differentiate the 2 functions in without multiplying them out.

)1( 210 xx

BUT dx

duv

910xdx

du

dx

dvx2

Let and10x 12 xu v

911 1010 xx 112x

Adding these gives the answer we want.

and dx

dvu

The Product Rule

dx

dvu

dx

duv

dx

dy

So, if uvy

where u and v are

both functions of x,

e.g. )1( 210 xxy10xu 12 xv

dx

dy )2(10 xx)1( 2 x 910x

910xdx

du x

dx

dv2

11911 21010 xxx 911 1012 xx

We need to simplify the answer:

Multiply out the brackets:Collect like terms:

The Product Rule

, but I’ve changed the

order.xx

dx

duv 2)1( 10

v is a function of a function but since the

derivative of the inner function is 1, we can

ignore it. However, we will meet more complicated functions of a function later!

The standard order is to have constants first, then powers of x and finally bracketed factors.

dx

dvu

dx

duv

dx

dy

102 )1( xxyLet

2xu

x2dx

du dx

dv 9)1(10 x

Now we can do (b) in the same way.

10)1( xvand

dx

dy 10)(2x 1x

The Product Rule

dx

dvu

dx

duv

dx

dy

102 )1( xxyLet

2xu

x2dx

du dx

dv 9)1(10 x

Now we can do (b) in the same way.

10)1( xvand

92 )(10 1xxdx

dy 10)(2x 1x

Don’t be tempted to try to multiply out. Think how many terms there will be!

There are common factors.

Tip: The cross ( multiply! ) acts as a reminder of the product

rule!

The Product Rule

How many ( x 1 ) factors are

common?

dx

dvu

dx

duv

dx

dy

102 )1( xxyLet

2xu

x2dx

du dx

dv 9)1(10 x

Now we can do (b) in the same way.

10)1( xvand

92 )(10 1xxdx

dy 10)(2x 1x

9)1( x2 xThe common factors.

x 92 )(10)( 1xx1x2 2 5

xx 5)1(

The Product Rule

9)1( x2 x

dx

dvu

dx

duv

dx

dy

102 )1( xxyLet

2xu

x2dx

du dx

dv 9)1(10 x

Now we can do (b) in the same way.

10)1( xvand

92 )(10 1xxdx

dy 10)(2x 1x

)16()1(2 9 xxx

xx 5)1(

92 )( 1xx2 5 x 10)( 1x2

The Product Rule

Otherwise use the product rule:

dx

dvu

dx

duv

dx

dy

If ,uvy

SUMMARY

where u and v are both functions

of x

To differentiate a product:

Check if it is possible to multiply out. If so, do it and differentiate each term.

The product rule says: • multiply the 2nd factor by the

derivative of the 1st.• Then add the 1st factor multiplied by

the derivative of the 2nd.

The Product Rule

N.B. You may, at first, find it difficult to simplify the answers to look the same as those given in textbooks. Don’t worry about this but keep trying as it gets easier with practice.

The Product Rule

Reminder:

xy sin3A function such as is a product, BUT we don’t need the product rule.

xdx

dyxy cos3sin3 e.g.

However, the product rule will work even though you shouldn’t use it

xvu sin3 and

N.B.

0dx

du

When we differentiate, a constant factor just “tags along” multiplying the answer to the 2nd factor.

The Product RuleExercis

eUse the product rule, where appropriate, to differentiate the following. Try to simplify your answers by removing common factors:

xexy 3

1.

2.

xxy sin43.

42 )2( xxy

4.

)2(2 xxy

The Product RuleSolutions:

1.

xexy 3

xx exexdx

dy 323 dx

dvu

dx

duv

dx

dy

23xdx

du xe

dx

dv

)3(2 xex x

3xu xev Let and

Remove common factors:

Notice the order within each term:

constants, powers of x, then

exponentials.

The Product Rule

xxxxdx

dycossin4 43

dx

dvu

dx

duv

dx

dy

34xdx

du x

dx

dvcos

)cossin4(3 xxxx

4xu xv sinLet and

Remove common factors:

2.

xxy sin4

3.

)2(2 xxy No need for the product rule: just multiply out.

322 xx 234 xx

dx

dy )34( xx

dx

dy

The Product Rule42 )2( xxy

4.

dx

dvu

dx

duv

dx

dy

xdx

du2 3)2(4 x

dx

dv

xxxx 2)2()2(2 3

2xu 4)2( xv Let and

Remove common factors:

xxx 32)2(2 3

324 )2(4)2(2 xxxxdx

dy

Did you notice that v was a function of a function?

Product Rule or Chain Rule?We can now differentiate all of the

following:

A simple function could be like any of the following:

We differentiate them term by term using the 4 rules for

173 2 xxy

xxy cos3sin2

21

xy xey 3

2

xxex xn cossin,, andThe multiplying constants just “tag along”.

simple functions,products andcompound functions ( functions of a function ).

Product Rule or Chain Rule?

Decide how you would differentiate each of the following ( but don’t do them ):

xy 2sin

xy 3sin

xxy sinxy sin2

(a)(b)(c)(d)

Product rule

Chain rule

This is a simple function

For products we use the product rule and for functions of a function we use the chain rule.

3)(sin xy Chain rule

Product Rule or Chain Rule?Exercis

eDecide with a partner how you would differentiate the following ( then do them if you need the practice ):1.

2sin xy

2.

2xey

3.

xxy sin 4.

)1( xey x 5.

xy 2sin

Write C for the Chain rule and P for the Product Rule

C

C

C or P

P

P

Product Rule or Chain Rule?

1.

xxy sin P xvxu sin

xdx

dv

dx

ducos1

xxxdx

dycossin

uyxu sin2 2coscos2 xu

du

dyx

dx

du

2cos2 xxdx

dy

2sin xy

2.

C

Solutions

dx

du

du

dy

dx

dy

dx

dvu

dx

duv

dx

dy

Product Rule or Chain Rule?

ueyxu 2

22 xu ee

du

dyx

dx

du

22 xex

dx

dy

3.

C2xey

xveu x 1

1dx

dve

dx

du x

)()1( xx exedx

dy

4.

)1( xey x P

xxedx

dy

dx

du

du

dy

dx

dy

dx

dvu

dx

duv

dx

dy

Product Rule or Chain Rule?

2sin uyxu

xudu

dyx

dx

dusin22cos

xxdx

dycossin2

5.

xy 2sin

xvxu sinsin

xdx

dvx

dx

ducoscos

xxxxdx

dycossincossin

xxdx

dycossin2

CEither

P

Or

dx

du

du

dy

dx

dy

dx

dvu

dx

duv

dx

dy