2.3 The Derivative
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2.3 The Derivative 1 General idea: introduce the derivative from geometric configuration as the slope of the line tangent to a curve at a given point. Method: we are going to find the tangent line as a limiting position for a secant line as points of intersection approach each other. T PQ Q line lim P T x
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T. x. 2.3 The Derivative. General idea: introduce the derivative from geometric configuration as the slope of the line tangent to a curve at a given point . - PowerPoint PPT Presentation
Transcript of 2.3 The Derivative
2.3 The Derivative
1
General idea: introduce the derivative from geometric configuration as the slope of the line tangent to a curve at a given point.
Method: we are going to find the tangent line as a limiting position for a secant line as points of intersection approach each other.
TPQQ
linelimP
T
x
2
Formalization: 1. Introduce coordinates for the point P: , where
2. Introduce differences between the coordinates of the point Q and point P: x and y. Now the coordinates of Q are where 3. The slope of the secant line PQ can be written as
),( 00 yx
),,( 00 yyxx ).( 00 xxfyy
).( 00 xfy
),( 00 yx
),( 00 yyxx
y
0x xx 0
x
x
y
3
Formalization (contd): 4. Combining 1 and 2 we can write y as
5. The statement QP can now be formalized as x0, and
)()( 00 xfxxf
))(,( 00 xxfxx
),( 00 yx
).()( 00 xfxxfy
x
T PQ
x
xfxxf
x
yPQ
xx
)()(limlim Slopelim 00
00PQ
4
Other notations:
Definition of the Derivative:
The formula obtained for the slope of the tangent line constitutes the definition of the derivative. We denote it as This is a derivative taken at one particular point but we can generalize it on any point in the domain of the function:
If the limit exists at some value of x, then the function is called differentiable at this point. If it exists at every point in an interval, the function is differentiable in the whole interval.
x
xfxxfxf
x
)()(lim)('
0
).(' 0xf,0x
.)(
')(' 0 dx
xdf
dx
dyyxf
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2.4 Differentiation by the Four-Step Process:
Step 1: In the function y=f(x), replace x by x+x and y by y+y:
Step2: Subtract y=f(x) from both sides:
Step 3: Divide both sides of the resulting expression by x:
Step 4: Obtain f’(x) by evaluating
).( xxfyy
).()( xfxxfy
x
xfxxf
x
y
)()(
x
yx
0
lim
6
Example: Find the derivative of
Solution: Step 1. for gives
Step 2. gives
Step 3. Dividing by x, we get
+ simplify the expression to get rid of x in the denominator…. Step 4.
xxy 3
)( xxfyy ) () () ( 3 f
)()( 3 xxxxyy
)()( xfxxfy
xxxxxxy 33 )()(
x
xxxxxx
x
y
33 )()(
13]1)(33[limlim)(' 222
00
xxxxxx
yxf
xx
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Example: Evaluate the derivative of at the points x=1, x=2
Solution: Step 1.
Step 2.
simplify the expression to get x as a factor…..
Step 3.
Step 4.
1
12
x
y
...)( xxfyy
1
1
1)(
122
xxx
y
11
222
xxx
xx
x
y
222200 )1(
2]
11
2[limlim)('
x
x
xxx
xx
x
yxf
xx
...)1(' f ...)2(' f
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Example: Find the derivative of
Solution: Step 1.
Step 2.
simplify the expression to get x as a factor…..
Step 3.
Step 4.
2 xy
...)( xxfyy
22 xxxy
22
1
xxxx
y
22
1
22
1limlim)('
00
xxxxx
yxf
xx
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HomeworkSection 2.4: 3,5,9,11,15,19,21,23.
