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Transcript of Today - The Product Rule Diff functions that are products Like © Christine Crisp Last lesson –...
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