ch.3 the derivative. differentiation given a curve y = f(x ) want to compute the slope of tangent at...
TRANSCRIPT
Ch.3 The Derivative
Differentiation
Given a curve y = f(x)
Want to compute the slope of tangent at some value x=a.
Let A=(a, f(a)) and
B=(a + x , f(a + x )) = (a + x , f(a) + y) where the change in y-value (y) is given by
y = f(a + x ) – f(a)
and B is a point on the curve y = f(x) close to A.
Therefore slope of chord
tangentchord the,0 As
)()(
ABxx
afxaf
x
yAB
Definition The derivative of f is defined by
This gives the slope of tangent at x = a
after setting x =a +x. We say that f is differentiable at a if this limit exists.
ax
afxfaf
ax
)()(lim)('
x
afxafaf
x
)()(lim)('
0
Thus the derivative of f at x is
Just set h = x and a=x.
We say that f is differentiable at x if the above limit exists.
This defines a new function f’(x) called the derivative of f(x) .
)(')()(
limlim00
xfh
xfhxf
x
y
dx
dyhx
Other notations for f’(x)
Examples
1. , show that
)(or or xDfdx
df
dx
dy
4 3 ) ( x x f y )('3 xfdx
dy
.expected as3y identicall is,)()(
AB, chord theof
slope theso linestraight a ofequation theis 43)( :Note
required. as,33lim3
lim
]43[]433[lim
]43[]4)(3[lim
)()(lim
lim
00
0
0
0
0
h
xfhxf
xxfyh
hh
xhxh
xhxh
xfhxfx
y
dx
dy
hh
h
h
h
x
2.
required. as,202)2(lim
2lim
)2(lim
)(lim
)()(lim
lim Now
).('2 that show )(
0
2
0
222
0
22
0
0
0
2
xxhxh
hxh
h
xhxhx
h
xhx
h
xfhxfx
y
dx
dy
xfxdx
dyxxf
h
h
h
h
h
x
3.
.
0000)1(
,,in terms)1([lim
])1([lim
)(lim
)()(lim)('
since
)('
)( If
1
2211
322211
0
22211
0
00
1
n
nn
nn
h
nnnnn
h
nn
hh
n
n
nx
xnnnx
etchhhxnnnx
h
xhhxnnhnxx
h
xhx
h
xfhxfxf
xfnxdx
dy
xxfy
Trigonometry identity:
Example
bababa sincoscossin)sin(
.cos1cos0sin
sinlimcos
1coslimsin
sincos
1cossinlim
sinsincoscossinlim
sin)sin(lim
)()(lim)(' :Proof
)('cossin)(
00
0
0
00
xxx
h
hx
h
hx
h
hx
h
hx
h
xhxhxh
xhx
h
xfhxfxf
xfxdx
dyxxfy
hh
h
h
hh
.001)10(cos
0sin1
)1(cos
sinsinlim
)1(cos
1coslim
)1(cos
)1)(cos1(coslim
1coslim:
0
22
0
00
h
h
h
h
hh
h
hh
hh
h
hNote
hh
hh
Example: Is f(x)=|x| differentiable at x=0?
Must consider
So the limit does not exists that is y=|x| is not differentiable at 0
0f1
0if1||But
.||
lim|0||0|
lim)0('
0at )()(
lim)('
00
0
hi
h
h
h
h
h
h
hf
xh
xfhxfxf
hh
h
Thus continuity does not imply differentiability.
Rate of Change
Recall that the slope of the tangent at a point measures the (instantaneous) rate of change of y with respect to x at that point.Example. Let s(t) be the distance of a car that has traveled at time t.
Speed v= rate of change of distance
)(' tsdt
dsv
Similarly
acceleration a = rate of change of speed
notationother
)()('' )2(2
2
tstsdt
sd
dt
ds
dt
d
dt
dva
ExampleA cylindrical tank holds 50 litres of water and can be drained from the bottom of the tank in 100 seconds. Find the rate of change of volume after 30 seconds given volume V of water in the tank after t seconds can be shown to be
Rate of change of volume
1000for 200
50)100
1(50)(2
2 tt
tt
tV
ond.litres/sec 0.8 rateat decreasing is volumeis,that
litres/sec 8.05
1120101
,20At
liter/sec. 101200
210
2
2
dt
dV
t
tt
dt
dV
Theorem
If f(x) is differentiable at a then f is continuous at x=a.
Proof. Assume f(x) differentiable at x=a.
Must show
But
0))()((lim
)()(lim
afxf
afxf
ax
ax
required. as
.00)('
)(lim)()(
lim))()((lim
)()()(
)()(
af
axax
afxfafxf
axax
afxfafxf
axaxax
Basic Differentiation Rules
Read Section 3.2 (or the same topic from other textbooks)
You should be able to use the differentiation rules/theorems to find the derivatives of functions