The Product Rule
uvvuy
uvy
In words:
“Keep the first, differentiate the second” +
“Keep the second, differentiate the first”
Examples: 1. Differentiate xxy cos
xxy cosxu xv cos1u xv sin
uvvuy
)1(cos)sin( xxxy
xxxy cossin
xxxy sincos
Examples: 2. Differentiate xxy 3sin2
xxy 3sin22xu xv 3sinxu 2 xv 3cos3
uvvuy
)2(3sin)3cos3(2 xxxxy
xxxxy 3sin23cos3 2 Now watch this.
Examples: 3. Differentiate xxy 3112 2
Try this using “words” xxy 3112 2
21
3112 2 xxy
)2(12231)3(312
112 2
1212 xxxxy
12314312
12321
21
2
xxx
xy
2
121
312
12318
312
123 2
x
xx
x
xy
x
xxxxy
312
15681443 22
x
xxy
312
115260 2
12314312
12321
21
2
xxx
xy
The Quotient Rule
2v
vuuvy
v
uy
In words:
“Keep the denominator, differentiate the numerator”
“Keep the numerator, differentiate the denominator”
–
Denominator 2
Examples: 1. Differentiate1
12
2
x
xy
12 xu
12 xv
xu 2
xv 2
2v
vuuvy
22
22
1
)2(1)2(1
x
xxxxy
22
22
1
112
x
xxx
22 1
22
x
x
22 1
4
x
x
Examples: 2. Differentiate 1
22
x
xy
1
)2(121
2)2(1
2
22 21
21
x
xxxxy
Try this using “words” 21
1
22
x
xy
1
12122
222 21
21
x
xxxy
1
12122
222 21
21
x
xxxy
11
2
1
12
2
2
2
2
2
21
21
x
x
x
x
x
y
11
212 2
2
22
21
x
x
xxy
1
1
1
222 2
1
xxy
23
1
22
x
y
Add a denominator here
Derivatives of New Functions
Definitions:x
xcos
1sec
xx
sin
1cosec
xx
tan
1cot
Reminder:
x
xx
cos
sintan
continue
xy sec
xy cosec
xy cot
Derivative of xtan xxdx
d 2sectan xy tan
Use the Quotient Rule now
Proof:
x
xy
cos
sin
x
xxxxy
2cos
)sin(sin)(coscos
x
xxy
2
22
cos
sincos
xy
2cos
1
xy 2sec
Derivatives of xsec
xxxdx
dtansecsec
x cosec, xcot and
xxxdx
dcot cosec cosec
xxdx
d coseccot 2
Prove these and keep with your
notes.Use chain
rule or quotient rule
Example: Given that xxxf tansin)( 2 show that 24
f
xxxf tansin)( 2
xx 22 secsin )(xf ))(cos(sin2 tan xxx
xxxf tansin)( 2
xx
22
cos
1sin
1
))(cos(sin2
cos
sin xx
x
x
x2tan x2sin2
4f
4tan2
4
sin2 2
12
2
12
2
Exponential and Logarithmic Functions
Reminder: xexf )( xxf ln)( and are inverse to each other.
They are perhaps the most important functions in the applications of calculus in the real world.
Alternative notation: xex exp as written becan
xx elog as written becan ln
Two very useful results: xe x ln
xex )ln(
Also: Practise changing from exp to log and vice-versa.
Learn these!
xey xy ln
xy
x
y
Derivatives of the Exponential and Logarithmic Functionsxey xy lnxey
xy
1
Proof of (ii)
(i) (ii)
xy lnxy elog
yex ye
dy
dx
yedx
dy 1
xdx
dy 1
Examples: 1. Differentiatexey sin
xey sin
xey x cossinxexy sincos
Use the Chain Rule
2. Differentiatexexy 4
xexy 4 Use the Product Rule
34 4xeexy xx
)4(3 xexy x
3. Differentiate 1ln 2 xy 1ln 2 xy
)2(1
12
xx
y
1
22
x
xy
Use the Chain Rule
4. Differentiatex
xy
ln
x
xy
ln
Use the Quotient Rule
2ln
1)1(ln
x
xxx
y
2ln
1ln
x
xy
Note: In general
)()( )( xfxf exfedx
d •
)(
)()(ln
xf
xfxf
dx
d •
Useful for reverse i.e. INTEGRATION
Higher Derivatives
Given that f is differentiable, if is also differentiable then its derivative is denoted by .
f f
The two notations are:
function1st
derivative2nd
derivative……
nth derivative
f ……
……
f f )(nf
dx
df2
2
dx
fdn
n
dx
fd
Example:
If , write down is first second and third derivatives and hence make a conjecture about its nth derivative.
xxey
xxey xx exe
dx
dy
xxxxx exeeexedx
yd2
2
2
xxxxx exeeexedx
yd32
3
3
n
n
dx
ydxx nexe
Conjecture: The nth derivative is n
n
dx
yd xx nexe
Rectilinear Motion
If displacement from the origin is a function of time I.e.
then
dt
dxv
)(tfx
v - velocity
2
2
dt
xd
dt
dva a - acceleration
Example: A body is moving in a straight line, so that after t seconds its displacement x metres from a fixed point O, is given by 3239 tttx
(a) Find the initial dislacement, velocity and acceleration of the body.
(b) Find the time at which the body is instantaneously at rest.
0t m 0)0()0(3)0(9 32 x
2369 ttdt
dxv m/s 9v
tdt
dva 66 2m/s 6a
Extreme Values of a Function
Understand the following terms:
• Critical Points
• Local Extreme Values
Local maximum
Local minimum
• End Point Extreme Values
End Point maximum
End Point minimum
See, MIA Mathematics 1, Pages 54 – 55
x
y
)(xfy
)(xfy
A
B
Consider maximum turning point A.
Notice, gradient of for x in the neighbourhood of A is negative.
)(xf
The Nature of Stationary Points
i.e. is negative)(xf
Similarly, gradient of for x in the neighbourhood of B is positive.
)(xf
i.e. is positive)(xf
Consider a curve and the corresponding gradient function
)(xfy )(xfy
The Nature of Stationary Points
Rule for Stationary Points
• and minimum turning point0)( xf 0)(xf
• and maximum turning point0)( xf 0)(xf
• and possibly a point of inflexion but must check using a table of signs
0)( xf 0)(xf
Example:
Consider 4)( xxf 34)( xxf
At S.P. 0)( xf
04 3 x0 x
212)( xxf
0)0( f
Now what does look like? 4)( xxf
x
y
Notice no Point of Inflexion.
Global Extreme Values
Understand the following terms:
• Global Extreme Values
Global maximum
Global minimum
See, MIA Mathematics 1, Pages 58 – 59
Example:
Find the coordinates and nature of the stationary point on the curve
.4)( xexf x
xexf x 4)(
4)( xexf
At S.P. 0)( xf
04 xe
4 xe
4ln x
4ln44ln ey4ln44 y
xexf )(
04)4(ln 4ln ef
)4ln44,4(ln is a
Minimum Turning Point
What does this curve look like?
xey x 4
x
y
x
y
Optimisation Problems
A sector of a circle with radius r cm has an area of 16 cm2.
Show that the perimeter P cm of the sector is given by
.16
2)(
rrrP
(a)
(b) Find the minimum value of P.
(a)2
sector area
2
length arc
rr
2
16
2 rr
l
2
)216
πr
πr(l
r rl
32
l
r
r
lrP 2now
rrP
322
rrP
162
1322)( rrrP(b)
At SP 0)( rP
2322)( rrP
2
322
r
032
22
r
2
322
r
322 2 r
162 r
4r
364)( rrP
3
64
r
34
64)4( P
01 r = 4 gives a minimum
stationary value of
4
32)4(2)4( P
cm 16
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