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Slopes and the Derivative
Mathematics 53
Institute of Mathematics (UP Diliman)
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For today
1 The Tangent Line
2 Definition of the Derivative
3 Differentiability
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For today
1 The Tangent Line
2 Definition of the Derivative
3 Differentiability
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Tangent Line to a Circle
Tangent line to a circle at point P
P
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Tangent Line to a Circle
Tangent line to a circle at point P
P
How about tangent line to a curve?
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Secant Line
y =f(x)
P
P(x0, f(x0)) - point on the graph off
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Secant Line
y =f(x)
P
Q
P(x0, f(x0)) - point on the graph off
Q(x1, f(x1)) - another pt on the graph off
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Secant Line
y =f(x)
P
Q
P(x0, f(x0))
Q(x1, f(x1))
P Q is a secant line (i.e. a line thatpasses through any two points on acurve).
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Secant Line
y =f(x)
P
Q
f(x0)
f(x1)
x0 x1
P(x0, f(x0))
Q(x1, f(x1))
P Q is a secant line (i.e. a line thatpasses through any two points on acurve).
What is the slope ofP Q?
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Secant Line
y =f(x)
P
Q
f(x0)
f(x1)
x0 x1
P(x0, f(x0))
Q(x1, f(x1))
P Q is a secant line (i.e. a line thatpasses through any two points on acurve).
mPQ
= f(x1) f(x0)
x1 x0
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Secant Line
y =f(x)
P
Q
f(x0)
f(x1)
x0 x1
P(x0, f(x0))
Q(x1, f(x1))
P Q is a secant line (i.e. a line thatpasses through any two points on acurve).
mPQ
= f(x1) f(x0)
x1 x0Let x= x1 x0
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Secant Line
y =f(x)
P
Q
f(x0)
f(x1)
x0 x1
P(x0, f(x0))
Q(x1, f(x1))
P Q is a secant line (i.e. a line thatpasses through any two points on acurve).
mPQ
= f(x1) f(x0)
x1 x0Let x= x1 x0
mPQ
= f(x0+ x) f(x0)
x
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The Tangent Line to the graph off at P
y =f(x)
P
Q
f(x0)
f(x1)
x0 x1
Let Q get closer and closer to P,i.e.
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The Tangent Line to the graph off at P
y =f(x)
P
Q
f(x0)
f(x1)
x0 x1
Let Q get closer and closer to P,i.e.
Q
P
x1 x0x 0
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The Tangent Line to the graph off at P
y =f(x)
P
Q
f(x0)
f(x1)
x0 x1
Let Q get closer and closer to P,i.e.
Q
P
x1 x0x 0
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The Tangent Line to the graph off at P
y =f(x)
P
Q
f(x0)
f(x1)
x0 x1
Let Q get closer and closer to P,i.e.
Q
P
x1 x0x 0
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 6 / 34
h h h f f P
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The Tangent Line to the graph off at P
y =f(x)
P
Q
f(x0)
f(x1)
x0x1
Let Q get closer and closer to P,i.e.
Q
P
x1 x0x 0
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 6 / 34
Th T Li h h f f P
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The Tangent Line to the graph off at P
y =f(x)
P
Let Q get closer and closer to P,i.e.
Q
P
x1 x0x 0
The lineP Q will approach the line .
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 6 / 34
Th T li h h f f ( f( ))
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The Tangent line to the graph off at(x0, f(x0))
y =f(x)
P
Slope of the line
.
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Th T t li t th h f f t ( f( ))
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The Tangent line to the graph off at(x0, f(x0))
y =f(x)
P
Slope of the line
m=
.
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Th T t li t th h f f t ( f( ))
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The Tangent line to the graph off at(x0, f(x0))
y =f(x)
P
Slope of the line
m= limx1x0
f(x1) f(x0)x1 x0
.
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Th T t li t th h f f t ( f( ))
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The Tangent line to the graph off at(x0, f(x0))
y =f(x)
P
Slope of the line
m= limx1x0
f(x1) f(x0)x1 x0
Recall: x=x1 x0
.
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 7 / 34
The Tangent line to the graph of f at (x f(x ))
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The Tangent line to the graph off at(x0, f(x0))
y =f(x)
P
Slope of the line
m= limx1x0
f(x1) f(x0)x1 x0
Recall: x=x1 x0
m=
.
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The Tangent line to the graph of f at (x f(x ))
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The Tangent line to the graph off at(x0, f(x0))
y =f(x)
P
Slope of the line
m= limx1x0
f(x1) f(x0)x1 x0
Recall: x=x1 x0
m= limx0
f(x0+ x) f(x0)x
.
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 7 / 34
The Tangent line to the graph of f at (x0 f(x0))
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The Tangent line to the graph off at(x0, f(x0))
y =f(x)
P
Slope of the line
m= limx1x0
f(x1) f(x0)x1 x0
Recall: x=x1 x0
m= limx0
f(x0+ x) f(x0)x
The line is called the tangent line to the graph ofy=f(x) at P.
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The Tangent Line
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The Tangent Line
Definition
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The Tangent Line
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The Tangent Line
Definition
If the function f is defined at x= x0, then the tangent line to the graphoffat the point P(x0, f(x0)) is the line
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The Tangent Line
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The Tangent Line
Definition
If the function f is defined at x= x0, then the tangent line to the graphoffat the point P(x0, f(x0)) is the line
1 passing through Pwhose slope is given by
m= limx0
f(x0+ x)
f(x0)
x ,provided that this limit exists.
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The Tangent Line
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The Tangent Line
Definition
If the function f is defined at x= x0, then the tangent line to the graphoffat the point P(x0, f(x0)) is the line
1 passing through Pwhose slope is given by
m= limx0
f(x0+ x)
f(x0)
x ,provided that this limit exists.
2 with equation x=x0 if
lim
x0
f(x0+ x) f(x0)
x
= +
or
and
limx0+
f(x0+ x) f(x0)x
= + or
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The Tangent Line
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The Tangent Line
Definition
If the function f is defined at x= x0, then the tangent line to the graphoffat the point P(x0, f(x0)) is the line
1 passing through Pwhose slope is given by
m= limx0
f(x0+ x)
f(x0)
x ,
provided that this limit exists.
2 with equation x=x0 if
lim
x0
f(x0+ x) f(x0)
x
= +
or
and
limx0+
f(x0+ x) f(x0)x
= + orOtherwise, there is no tangent line to the graph off at P.
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 8 / 34
The Tangent Line
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The Tangent Line
Remarks
1 The slope of the tangent line to the graph off at Pgives us an ideaon the flatness or steepness of the graph off at Pand whetherthe graph offrises or falls at P.
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The Tangent Line
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g
Remarks
1 The slope of the tangent line to the graph off at Pgives us an ideaon the flatness or steepness of the graph off at Pand whetherthe graph offrises or falls at P.
0
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The Tangent Line
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g
Remarks
1 The slope of the tangent line to the graph off at Pgives us an ideaon the flatness or steepness of the graph off at Pand whetherthe graph offrises or falls at P.
2 The tangent line to the graph of a function may intersect the graph at
points other than the point of tangency.
0
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The Tangent Line
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g
Remarks
1 The slope of the tangent line to the graph off at Pgives us an ideaon the flatness or steepness of the graph off at Pand whetherthe graph offrises or falls at P.
2 The tangent line to the graph of a function may intersect the graph at
points other than the point of tangency.
0
P
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The Normal Line
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Definition
The line normal to the graph offat the point P is the line perpendicularto the tangent line at P.
P
0
NL
TL
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E l
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Example
Give equations of the tangent and normal lines to the graph off(x) = 1
x
at x= 1.
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E l
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Example
Give equations of the tangent and normal lines to the graph off(x) = 1
x
at x= 1.Solution.
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Example
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Example
Give equations of the tangent and normal lines to the graph off(x) = 1
x
at x= 1.Solution.
Recall: Point-Slope Form of a Line : y y0=m(x x0)
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Example
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Example
Give equations of the tangent and normal lines to the graph off(x) = 1
x
at x= 1.Solution.
Recall: Point-Slope Form of a Line : y y0=m(x x0)f(1) = 1 so the point is (1, 1)
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Example
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Example
Give equations of the tangent and normal lines to the graph off(x) = 1
x
at x= 1.Solution.
Recall: Point-Slope Form of a Line : y y0=m(x x0)f(1) = 1 so the point is (1, 1)m
TL
= slope of the tangent line to the graph off(x) at (1, 1)
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Example
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Example
Give equations of the tangent and normal lines to the graph off(x) = 1
x
at x= 1.Solution.
Recall: Point-Slope Form of a Line : y y0=m(x x0)f(1) = 1 so the point is (1, 1)m
TL
= slope of the tangent line to the graph off(x) at (1, 1)
mTL
= limx0
f(1 + x) f(1)x
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 11 / 34
Example
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Example
Give equations of the tangent and normal lines to the graph off(x) = 1
x
at x= 1.Solution.
Recall: Point-Slope Form of a Line : y y0=m(x x0)f(1) = 1 so the point is (1, 1)m
TL
= slope of the tangent line to the graph off(x) at (1, 1)
mTL
= limx0
f(1 + x) f(1)x
= limx0
11+x 1
x
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Example
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Example
Give equations of the tangent and normal lines to the graph off(x) = 1
x
at x= 1.Solution.
Recall: Point-Slope Form of a Line : y y0=m(x x0)f(1) = 1 so the point is (1, 1)m
TL
= slope of the tangent line to the graph off(x) at (1, 1)
mTL
= limx0
f(1 + x) f(1)x
= limx0
11+x 1
x
= limx0
1 (1 + x)x(1 + x)
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= limx0
xx(1 + x)
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= limx0
xx(1 + x)
= limx0
11 + x
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= limx0
xx(1 + x)
= limx0
11 + x
mTL
=
1
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= limx0
xx(1 + x)
= limx0
11 + x
mTL
=
1
Tangent Line: y 1 = (x 1)
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= limx0
xx(1 + x)
= limx0
11 + x
mTL
=
1
Tangent Line: y 1 = (x 1)
mTL
= 1, so the slope of the normal line is 1.
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= limx0
xx(1 + x)
= limx0
11 + x
mTL
=
1
Tangent Line: y 1 = (x 1)
mTL
= 1, so the slope of the normal line is 1.
Normal Line: y 1 = (x 1)
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Example
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Find the slope of the tangent line to the graph ofg(x) =x2 + 2 at x= 1and at x= 2.
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Example
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Find the slope of the tangent line to the graph ofg(x) =x2 + 2 at x= 1and at x= 2.
Solution. Atx= 1 :
mTL
= limx0
g(1 + x) g(1)x
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 13 / 34
Example
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Find the slope of the tangent line to the graph ofg(x) =x2 + 2 at x= 1and at x= 2.
Solution. Atx= 1 :
mTL
= limx0
g(1 + x) g(1)x
= limx0
[(1 + x)2 + 2]
(12 + 2)
x
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 13 / 34
Example
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Find the slope of the tangent line to the graph ofg(x) =x2 + 2 at x= 1and at x= 2.
Solution. Atx= 1 :
mTL
= limx0
g(1 + x) g(1)x
= limx0
[(1 + x)2 + 2]
(12 + 2)
x
= limx0
2x+ (x)2
x
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 13 / 34
Example
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Find the slope of the tangent line to the graph ofg(x) =x2 + 2 at x= 1and at x= 2.
Solution. Atx= 1 :
mTL
= limx0
g(1 + x) g(1)x
= limx0
[(1 + x)2 + 2]
(12 + 2)
x
= limx0
2x+ (x)2
x
= limx0
x(2 + x)
x
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Example
2
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Find the slope of the tangent line to the graph ofg(x) =x2 + 2 at x= 1and at x= 2.
Solution. Atx= 1 :
mTL
= limx0
g(1 + x) g(1)x
= limx0
[(1 + x)2 + 2]
(12 + 2)
x
= limx0
2x+ (x)2
x
= limx0
x(2 + x)
x= lim
x0(2 + x)
= 2.
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 13 / 34
Atx= 2 :
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Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 14 / 34
Atx= 2 :
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mTL
= limx0
g(2 + x) g(2)x
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Atx= 2 :
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mTL
= limx0
g(2 + x) g(2)x
= limx0
[(2 + x)2 + 2] (22 + 2)x
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 14 / 34
Atx= 2 :
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mTL
= limx0
g(2 + x) g(2)x
= limx0
[(2 + x)2 + 2] (22 + 2)x
= limx0
[4 + 4x+ (x)2 + 2] 6x
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 14 / 34
Atx= 2 :
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mTL
= limx0
g(2 + x) g(2)x
= limx0
[(2 + x)2 + 2] (22 + 2)x
= limx0
[4 + 4x+ (x)2 + 2] 6x
= limx0
4x+ (x)2x
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 14 / 34
Atx= 2 :
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mTL
= limx0
g(2 + x) g(2)x
= limx0
[(2 + x)2 + 2] (22 + 2)x
= limx0
[4 + 4x+ (x)2 + 2] 6x
= limx0
4x+ (x)2x
= limx0
x(4 + x)
x
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 14 / 34
Atx= 2 :
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mTL
= limx0
g(2 + x) g(2)x
= limx0
[(2 + x)2 + 2] (22 + 2)x
= limx0
[4 + 4x+ (x)2 + 2] 6x
= limx0
4x+ (x)2x
= limx0
x(4 + x)
x= lim
x0(4 + x)
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 14 / 34
Atx= 2 :
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mTL
= limx0
g(2 + x) g(2)x
= limx0
[(2 + x)2 + 2] (22 + 2)x
= limx0
[4 + 4x+ (x)2 + 2] 6x
= limx0
4x+ (x)2x
= limx0
x(4 + x)
x= lim
x0
(4 + x)
= 4.
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 14 / 34
As a matter of fact, we can compute for the slope of the tangent line tog(x) for any value ofx.
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( )
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 15 / 34
As a matter of fact, we can compute for the slope of the tangent line tog(x) for any value ofx.
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mTL = limx0
g(x+ x)
g(x)
x
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 15 / 34
As a matter of fact, we can compute for the slope of the tangent line tog(x) for any value ofx.
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mTL = limx0
g(x+ x)
g(x)
x
= limx0
[(x+ x)2 + 2] (x2 + 2)x
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 15 / 34
As a matter of fact, we can compute for the slope of the tangent line tog(x) for any value ofx.
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mTL
= limx0
g(x+ x)
g(x)
x
= limx0
[(x+ x)2 + 2] (x2 + 2)x
= lim
x
0
[x2 + 2xx+ (x)2 + 2] (x2 + 2)
x
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 15 / 34
As a matter of fact, we can compute for the slope of the tangent line tog(x) for any value ofx.
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mTL
= limx0
g(x+ x)
g(x)
x
= limx0
[(x+ x)2 + 2] (x2 + 2)x
= lim
x
0
[x2 + 2xx+ (x)2 + 2] (x2 + 2)
x= lim
x0
2xx+ (x)2
x
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 15 / 34
As a matter of fact, we can compute for the slope of the tangent line tog(x) for any value ofx.
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mTL
= limx0
g(x+ x)
g(x)
x
= limx0
[(x+ x)2 + 2] (x2 + 2)x
= lim
x
0
[x2 + 2xx+ (x)2 + 2] (x2 + 2)
x= lim
x0
2xx+ (x)2
x= lim
x0(2x+ x)
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 15 / 34
As a matter of fact, we can compute for the slope of the tangent line tog(x) for any value ofx.
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mTL
= limx0
g(x+ x) g(x)x
= limx0
[(x+ x)2 + 2] (x2 + 2)x
= limx0
[x2 + 2xx+ (x)2 + 2] (x2 + 2)
x= lim
x0
2xx+ (x)2
x= lim
x0(2x+ x)
= 2x.
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 15 / 34
As a matter of fact, we can compute for the slope of the tangent line tog(x) for any value ofx.
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mTL
= limx0
g(x+ x) g(x)x
= limx0
[(x+ x)2 + 2] (x2 + 2)x
= limx0
[x2 + 2xx+ (x)2 + 2] (x2 + 2)
x= lim
x0
2xx+ (x)2
x= lim
x0(2x+ x)
= 2x.
Atx= 1:
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 15 / 34
As a matter of fact, we can compute for the slope of the tangent line tog(x) for any value ofx.
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mTL
= limx0
g(x+ x) g(x)x
= limx0
[(x+ x)2 + 2] (x2 + 2)x
= limx0
[x2 + 2xx+ (x)2 + 2] (x2 + 2)
x= lim
x0
2xx+ (x)2
x= lim
x0(2x+ x)
= 2x.
Atx= 1: mTL= 2 1 = 2
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 15 / 34
As a matter of fact, we can compute for the slope of the tangent line tog(x) for any value ofx.
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mTL
= limx0
g(x+ x) g(x)x
= limx0
[(x+ x)2 + 2] (x2 + 2)x
= limx0
[x2 + 2xx+ (x)2 + 2] (x2 + 2)
x
= limx0
2xx+ (x)2
x= lim
x0(2x+ x)
= 2x.
Atx= 1: mTL= 2 1 = 2Atx= 2:
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 15 / 34
As a matter of fact, we can compute for the slope of the tangent line tog(x) for any value ofx.
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mTL
= limx0
g(x+ x) g(x)x
= limx0
[(x+ x)2 + 2] (x2 + 2)x
= limx0
[x2 + 2xx+ (x)2 + 2] (x2 + 2)
x
= limx0
2xx+ (x)2
x= lim
x0(2x+ x)
= 2x.
Atx= 1: mTL= 2 1 = 2Atx= 2: mTL= 2 2 = 4
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 15 / 34
For today
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1 The Tangent Line
2 Definition of the Derivative
3 Differentiability
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Definition of the Derivative
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In the previous example, we computed a function that gives the slope ofthe tangent line to the graph of a function f(x) at any value ofx. Thisfunction is actually called the derivative off(x).
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Definition of the Derivative
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In the previous example, we computed a function that gives the slope ofthe tangent line to the graph of a function f(x) at any value ofx. Thisfunction is actually called the derivative off(x).
Definition
The derivative of a function f(x), denoted by f
(x), is the function
f(x) = limx0
f(x+ x) f(x)x
.
It is defined at all pointsx
in the domain off
where the limit exists.
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 17 / 34
Definition of the Derivative
Remarks
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Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 18 / 34
Definition of the Derivative
Remarks
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1
There may be points xo dom fat which f
(x0) does not exist. So,dom f dom f.
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 18 / 34
Definition of the Derivative
Remarks
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1
There may be points xo dom fat which f
(x0) does not exist. So,dom f dom f.2 The definition tells us that f(x0) is the slope of the tangent line to
the graph of the function at point P(x0, f(x0)).
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Definition of the Derivative
Remarks
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1
There may be points xo dom fat which f
(x0) does not exist. So,dom f dom f.2 The definition tells us that f(x0) is the slope of the tangent line to
the graph of the function at point P(x0, f(x0)).
3 To get the derivative off at x=x0, we use
f(x0) = limx0
f(x0+ x) f(x0)x
.
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 18 / 34
Definition of the Derivative
Remarks
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1
There may be points xo dom fat which f
(x0) does not exist. So,dom f dom f.2 The definition tells us that f(x0) is the slope of the tangent line to
the graph of the function at point P(x0, f(x0)).
3 To get the derivative off at x=x0, we use
f(x0) = limx0
f(x0+ x) f(x0)x
.
Alternatively,
f(x0) = limxx0
f(x) f(x0)x x0
.
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 18 / 34
Definition of the Derivative
Remarks
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1
There may be points xo dom fat which f
(x0) does not exist. So,dom f dom f.2 The definition tells us that f(x0) is the slope of the tangent line to
the graph of the function at point P(x0, f(x0)).
3 To get the derivative off at x=x0, we use
f(x0) = limx0
f(x0+ x) f(x0)x
.
Alternatively,
f(x0) = limxx0
f(x) f(x0)x x0
.
4 Other notations: y ify=f(x), dy
dx ,
d
dx[f(x)] , Dx[f(x)].
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 18 / 34
Definition of the Derivative
Remarks
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1
There may be points xo dom fat which f
(x0) does not exist. So,dom f dom f.2 The definition tells us that f(x0) is the slope of the tangent line to
the graph of the function at point P(x0, f(x0)).
3 To get the derivative off at x=x0, we use
f(x0) = limx0
f(x0+ x) f(x0)x
.
Alternatively,
f(x0) = limxx0
f(x) f(x0)x x0
.
4 Other notations: y ify=f(x), dy
dx ,
d
dx[f(x)] , Dx[f(x)].
5 The process of computing the derivative is called differentiation.
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 18 / 34
Definition of the Derivative
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Example
Find the derivative off(x) = x.
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Definition of the Derivative
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Example
Find the derivative off(x) = x.
Solution. Using the definition of the derivative we have
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Definition of the Derivative
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Example
Find the derivative off(x) = x.
Solution. Using the definition of the derivative we have
f
(x) = limx0
x+ x
x
x
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Definition of the Derivative
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Example
Find the derivative off(x) = x.
Solution. Using the definition of the derivative we have
f
(x) = limx0
x+ x
x
x
= limx0
x+ xx
x
x+ x+
x
x+ x+
x
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 19 / 34
Definition of the Derivative
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Example
Find the derivative off(x) = x.
Solution. Using the definition of the derivative we have
f
(x) = limx0
x+ x
x
x
= limx0
x+ xx
x
x+ x+
x
x+ x+
x
= limx0
(x+ x)
x
x(x+ x+ x)
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 19 / 34
1
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= limx0
1
x+ x+ x
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 20 / 34
1
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= limx0
1
x+ x+ x=
1
2
x.
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 20 / 34
1
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= limx0
1
x+ x+ x=
1
2
x.
f
(x) =
1
2x or Dx[x ] = 1
2x .
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 20 / 34
1
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= limx0
1
x+ x+ x=
1
2
x.
f
(x) =
1
2x or Dx[x ] = 1
2x .Note:
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 20 / 34
1
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= limx0
1
x+ x+ x=
1
2
x.
f
(x) =
1
2x or Dx[x ] = 1
2x .Note:
dom f = [0, +)
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 20 / 34
1
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= limx0
1
x+ x+ x=
1
2
x.
f
(x) =
1
2x or Dx[x ] = 1
2x .Note:
dom f = [0, +)dom f = (0, +
)
Instit te of Mathematics (UP Diliman) Slo es and the De i ati e Mathematics 53 20 / 34
Example
Find the derivative off(x) = 3
x at x= 0.
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I tit t f M th ti (UP Dili ) Sl d th D i ti M th ti 53 21 / 34
Example
Find the derivative off(x) = 3
x at x= 0.
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Solution.
I tit t f M th ti (UP Dili ) Sl d th D i ti M th ti 53 21 / 34
Example
Find the derivative off(x) = 3
x at x= 0.
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Solution.Using the alternative definition:
f(0) = limx0
3
x 30x 0
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Example
Find the derivative off(x) = 3
x at x= 0.
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Solution.Using the alternative definition:
f(0) = limx0
3
x 30x 0
= limx0
3
xx
I tit t f M th ti (UP Dili ) Sl d th D i ti M th ti 53 21 / 34
Example
Find the derivative off(x) = 3
x at x= 0.
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Solution.Using the alternative definition:
f(0) = limx0
3
x 30x 0
= limx0
3
xx
= limx0
1
x23
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 21 / 34
Example
Find the derivative off(x) = 3
x at x= 0.
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Solution.Using the alternative definition:
f(0) = limx0
3
x 30x 0
= limx0
3
xx
= limx0
1
x23
=
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 21 / 34
Example
Find the derivative off(x) = 3
x at x= 0.
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Solution.Using the alternative definition:
f(0) = limx0
3
x 30x 0
= limx0
3
xx
= limx0
1
x23
= 1
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 21 / 34
Example
Find the derivative off(x) = 3
x at x= 0.
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Solution.Using the alternative definition:
f(0) = limx0
3
x 30x 0
= limx0
3
xx
= limx0
1
x23
= 1
0+
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 21 / 34
Example
Find the derivative off(x) = 3
x at x= 0.
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Solution.Using the alternative definition:
f(0) = limx0
3
x 30x 0
= limx0
3
xx
= limx0
1
x23
= 1
0+
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 21 / 34
For today
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1 The Tangent Line
2 Definition of the Derivative
3 Differentiability
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 22 / 34
Differentiability
Definitions
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Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 23 / 34
Differentiability
Definitions
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1 If the derivative f(x0) of the function f at x= x0 exists, then f issaid to be differentiable at x= x0.
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 23 / 34
Differentiability
Definitions
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1 If the derivative f(x0) of the function f at x= x0 exists, then f issaid to be differentiable at x= x0.
2 The function f is differentiable on (a, b) iff is differentiable at
every real number in (a, b
).
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 23 / 34
Differentiability
Definitions
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1 If the derivative f(x0) of the function f at x= x0 exists, then f issaid to be differentiable at x= x0.
2 The function f is differentiable on (a, b) iff is differentiable atevery real number in (a, b).
3 The function f is differentiable everywhere if it is differentiable atevery real number.
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 23 / 34
Differentiability
Definitions
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1 If the derivative f(x0) of the function f at x= x0 exists, then f issaid to be differentiable at x= x0.
2 The function f is differentiable on (a, b) iff is differentiable atevery real number in (a, b).
3 The function f is differentiable everywhere if it is differentiable atevery real number.
RemarkThe function f(x) is differentiable at x= x0 if and only ifx0dom f.
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Example
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ExampleWe know that Dx[
x ] =
1
2
x.
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Example
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Example
We know that Dx[
x ] = 1
2
x.
= f(x) = x is differentiable at any positive real number x
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Example
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Example
We know that Dx[
x ] = 1
2
x.
= f(x) = x is differentiable at any positive real number x
= f is not differentiable at x= 0.
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 24 / 34
Differentiability
Definitions
Let the function f(x) be defined at x= x0.
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f ( )
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 25 / 34
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Differentiability
Definitions
Let the function f(x) be defined at x= x0.
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1 The derivative from the left off(x) at x=x0, denoted by f
(x0),
is given by
f
(x0) = lim
xx
0
f(x) f(x0)
x x0.
2 The derivative from the right off(x) at x= x0, denoted byf+(x0), is given by
f+(x0) = limxx+0
f(x) f(x0)x x0 .
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Differentiability
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Remark
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Differentiability
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Remark
1 The derivative from the left [right] is also referred to as the left-handderivative [right-hand derivative], or simply left derivative [rightderivative].
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Differentiability
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Remark
1 The derivative from the left [right] is also referred to as the left-handderivative [right-hand derivative], or simply left derivative [rightderivative].
2 The function f is differentiable at x=x0 if and only iff
(x0) and
f+(x0) exist and
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 26 / 34
Differentiability
k
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Remark
1 The derivative from the left [right] is also referred to as the left-handderivative [right-hand derivative], or simply left derivative [rightderivative].
2 The function f is differentiable at x=x0 if and only iff
(x0) and
f+(x0) exist and
f
(x0) =f
+(x0) =f(x0).
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 26 / 34
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Example
Determine iff(x) = |x| =
x , x 0x , x
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Solution.
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Example
Determine iff(x) = |x| =
x , x 0x , x
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Solution. We compute for the derivatives from the left and from the rightat x= 0.
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Example
Determine iff(x) = |x| =
x , x 0x , x
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Solution. We compute for the derivatives from the left and from the rightat x= 0.
f
(0)
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Example
Determine iff(x) = |x| =
x , x 0x , x
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Solution. We compute for the derivatives from the left and from the rightat x= 0.
f
(0) = limx0
f(x) f(0)x
0
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Example
Determine iff(x) = |x| =
x , x 0x , x
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Solution. We compute for the derivatives from the left and from the rightat x= 0.
f
(0) = limx0
f(x) f(0)x
0
= limx0
x 0x
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Example
Determine iff(x) = |x| =
x , x 0x , x
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Solution. We compute for the derivatives from the left and from the rightat x= 0.
f
(0) = limx0
f(x) f(0)x
0
= limx0
x 0x
= 1
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Example
Determine iff(x) = |x| =
x , x 0x , x
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Solution. We compute for the derivatives from the left and from the rightat x= 0.
f
(0) = limx0
f(x) f(0)x
0
= limx0
x 0x
= 1
f+(0)
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Example
Determine iff(x) = |x| =
x , x 0x , x
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Solution. We compute for the derivatives from the left and from the rightat x= 0.
f
(0) = limx0
f(x) f(0)x
0
= limx0
x 0x
= 1
f+(0) = limx0+
f(x) f(0)x
0
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Example
Determine iff(x) = |x| =
x , x 0x , x
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Solution. We compute for the derivatives from the left and from the rightat x= 0.
f
(0) = limx0
f(x) f(0)x
0
= limx0
x 0x
= 1
f+(0) = limx0+
f(x) f(0)x
0
= limx0+
x 0x
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 27 / 34
Example
Determine iff(x) = |x| =
x , x 0x , x
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Solution. We compute for the derivatives from the left and from the rightat x= 0.
f
(0) = limx0
f(x) f(0)x
0
= limx0
x 0x
= 1
f+(0) = limx0+
f(x) f(0)x
0
= limx0+
x 0x
= 1
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Example
Determine iff(x) = |x| =
x , x 0x , x
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Solution. We compute for the derivatives from the left and from the rightat x= 0.
f
(0) = limx0
f(x) f(0)x
0
= limx0
x 0x
= 1
f+(0) = limx0+
f(x) f(0)x
0
= limx0+
x 0x
= 1
=
f
(0)
=f+(0)
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 27 / 34
Example
Determine iff(x) = |x| =
x , x 0x , x
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Solution. We compute for the derivatives from the left and from the rightat x= 0.
f
(0) = limx0
f(x) f(0)x
0
= limx0
x 0x
= 1
f+(0) = limx0+
f(x) f(0)x
0
= limx0+
x 0x
= 1
=
f
(0)
=f+(0)
f(x) is not differentiable at x= 0.
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 27 / 34
Example
Determine iff(x) =
12x
2 + 24 , x
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Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 28 / 34
Example
Determine iff(x) =
12x
2 + 24 , x
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We compute for the derivatives from the left and from the right at x= 4.
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Example
Determine iff(x) =
12x
2 + 24 , x
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We compute for the derivatives from the left and from the right at x= 4.
f
(4)
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Example
Determine iff(x) =
12x
2 + 24 , x
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We compute for the derivatives from the left and from the right at x= 4.
f
(4) = limx4
( 12
x2 + 24) 32x 4
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Example
Determine iff(x) =
12x
2 + 24 , x
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We compute for the derivatives from the left and from the right at x 4.
f
(4) = limx4
( 12
x2 + 24) 32x 4
= limx4
1
2x2
8x 4
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 28 / 34
Example
Determine iff(x) =
12x
2 + 24 , x
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We compute for the derivatives from the left and from the right at x 4.
f
(4) = limx4
( 12
x2 + 24) 32x 4
= limx4
1
2
x2
8x 4= lim
x4
1
2
(x+ 4)(x 4)x 4
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 28 / 34
Example
Determine iff(x) =
12x
2 + 24 , x
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p g
f
(4) = limx4
( 12
x2 + 24) 32x 4
= limx4
1
2
x2
8x 4= lim
x4
1
2
(x+ 4)(x 4)x 4
= 4
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 28 / 34
Example
Determine iff(x) =
12x
2 + 24 , x
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p g
f
(4) = limx4
( 12
x2 + 24) 32x 4
= limx4
1
2
x2
8
x 4= lim
x4
1
2
(x+ 4)(x 4)x 4
= 4
f+(4)
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 28 / 34
Example
Determine iff(x) =
12x
2 + 24 , x
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p g
f
(4) = limx4
( 12
x2 + 24) 32x 4
= limx4
1
2
x2
8
x 4= lim
x4
1
2
(x+ 4)(x 4)x 4
= 4
f+(4) = limx4+
16
x 32x 4
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 28 / 34
Example
Determine iff(x) =
12x
2 + 24 , x
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f
(4) = limx4
( 12
x2 + 24) 32x 4
= limx4
1
2
x2
8
x 4= lim
x4
1
2
(x+ 4)(x 4)x 4
= 4
f+(4) = limx4+
16
x 32x 4
= limx4+ 16
x
2
x 4
x+ 2x+ 2
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 28 / 34
Example
Determine iff(x) =
12x
2 + 24 , x
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f
(4) = limx4
( 12
x2 + 24) 32x 4
= limx4
1
2
x2
8
x 4= lim
x4
1
2
(x+ 4)(x 4)x 4
= 4
f+(4) = limx4+
16
x 32x 4
= limx4+ 16
x
2
x 4
x+ 2x+ 2
= limx4+
16 x 4
(x 4)x+ 2
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 28 / 34
Example
Determine iff(x) =
12x
2 + 24 , x
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f
(4) = limx4
( 12
x2 + 24) 32x 4
= limx4
1
2
x2
8
x 4= lim
x4
1
2
(x+ 4)(x 4)x 4
= 4
f+(4) = limx4+
16
x 32x 4
= limx4+ 16
x
2
x 4
x+ 2x+ 2
= limx4+
16 x 4
(x 4)x+ 2= 4
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 28 / 34
Example
Determine iff(x) =
12x
2 + 24 , x
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f
(4) = limx4
( 12
x2 + 24) 32x 4
= limx4
1
2
x2
8
x 4= lim
x4
1
2
(x+ 4)(x 4)x 4
= 4
f+(4) = limx4+
16
x 32x 4
= limx4+ 16
x
2
x 4
x+ 2x+ 2
= limx4+
16 x 4
(x 4)x+ 2= 4
f
(4) =f+(4) = 4
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 28 / 34
Example
Determine iff(x) =
12x
2 + 24 , x
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f
(4) = limx4
( 12
x2 + 24) 32x 4
= limx4
1
2
x2
8
x 4= lim
x4
1
2
(x+ 4)(x 4)x 4
= 4
f+(4) = limx4+
16
x 32x 4
= limx4+ 16
x
2
x 4
x+ 2x+ 2
= limx4+
16 x 4
(x 4)x+ 2= 4
f
(4) =f+(4) = 4 = f(x) is differentiable at x= 4
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 28 / 34
Example
Determine iff(x) =
12x
2 + 24 , x
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f
(4) = limx4
( 12
x2 + 24) 32x 4
= limx4
1
2
x2
8
x 4= lim
x4
1
2
(x+ 4)(x 4)x 4
= 4
f+(4) = limx4+
16
x 32x 4
= limx4+ 16
x
2
x 4
x+ 2x+ 2
= limx4+
16 x 4
(x 4)x+ 2= 4
f
(4) =f+(4) = 4 = f(x) is differentiable at x= 4 and f(4) = 4.
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 28 / 34
Differentiability and Continuity
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Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 29 / 34
Differentiability and Continuity
Theorem
Iff is differentiable atx= x0, then f is continuous atx=x0.
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Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 29 / 34
Differentiability and Continuity
Theorem
Iff is differentiable atx= x0, then f is continuous atx=x0.
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Remarks
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 29 / 34
Differentiability and Continuity
Theorem
Iff is differentiable atx= x0, then f is continuous atx=x0.
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Remarks
Iff is discontinuous at x= x0, then f is not differentiable at x=x0.
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 29 / 34
Differentiability and Continuity
Theorem
Iff is differentiable atx= x0, then f is continuous atx=x0.
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Remarks
Iff is discontinuous at x= x0, then f is not differentiable at x=x0.
Iff is continuous at x= x0, it does not mean that f is differentiableat x=x0.
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 29 / 34
Differentiability and Continuity
Theorem
Iff is differentiable atx= x0, then f is continuous atx=x0.
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Remarks
Iff is discontinuous at x= x0, then f is not differentiable at x=x0.
Iff is continuous at x= x0, it does not mean that f is differentiableat x=x0.
Iff is not differentiable at x= x0, it does not mean that f is notcontinuous at x= x0.
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 29 / 34
Example
Consider the function f(x) = |x|.
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Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 30 / 34
Example
Consider the function f(x) = |x|.
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f is continuous at x= 0
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 30 / 34
Example
Consider the function f(x) = |x|.
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f is continuous at x= 0
f is not differentiable at x= 0
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 30 / 34
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Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 31 / 34
Geometrically, iff is differentiable at x=x0, then the graph off hastangent line at x=x0.
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Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 31 / 34
Geometrically, iff is differentiable at x=x0, then the graph off hastangent line at x=x0.
Remarks
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Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 31 / 34
Geometrically, iff is differentiable at x=x0, then the graph off hastangent line at x=x0.
Remarks
A function f is not differentiable at x=x0 if one of the following is true:
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Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 31 / 34
Geometrically, iff is differentiable at x=x0, then the graph off hastangent line at x=x0.
Remarks
A function f is not differentiable at x=x0 if one of the following is true:f is discontinuous at x= x0
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Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 31 / 34
Geometrically, iff is differentiable at x=x0, then the graph off hastangent line at x=x0.
Remarks
A function f is not differentiable at x=x0 if one of the following is true:f is discontinuous at x= x0
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x0
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 31 / 34
Remarks
the graph offhas a vertical tangent line at x=x0
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Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 32 / 34
Remarks
the graph offhas a vertical tangent line at x=x0
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x0
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 32 / 34
Remarks
the graph offhas a vertical tangent line at x=x0
the graph offhas no well-defined tangent line at x=x0, i.e., the
graph offhas a corner or cusp at x= x0.
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x0
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 32 / 34
Remarks
the graph offhas a vertical tangent line at x=x0
the graph offhas no well-defined tangent line at x=x0, i.e., the
graph offhas a corner or cusp at x= x0.
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x0 x0
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 32 / 34
* * * The End * * *
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Next Meeting:
Differentiation Rules
The Chain Rule
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 33 / 34
Exercise
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Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 34 / 34
Exercise
Determine if
f(x) =
x2 , x < 1
1 2x , x 1is differentiable at x0= 1.
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Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 34 / 34
Exercise
Determine if
f(x) =
x2 , x < 1
1 2x , x 1is differentiable at x0= 1.Solution.
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Solution.
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 34 / 34
Exercise
Determine if
f(x) =
x2 , x < 1
1 2x , x 1is differentiable at x0= 1.Solution.
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f
(1) = limx1
x2 1x+ 1
= limx1
(x+ 1)(x 1)x+ 1
= 2
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 34 / 34
Exercise
Determine if
f(x) =
x2 , x < 1
1 2x , x 1is differentiable at x0= 1.Solution.
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f
(1) = limx1
x2 1x+ 1
= limx1
(x+ 1)(x 1)x+ 1
= 2
f+(
1) = lim
x3
+
(1 2x) 1
x+ 1= lim
x3+
2(x+ 1)x+ 1
= 2
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 34 / 34
Exercise
Determine if
f(x) =
x2 , x < 1
1 2x , x 1is differentiable at x0= 1.Solution.
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f
(1) = limx1
x2 1x+ 1
= limx1
(x+ 1)(x 1)x+ 1
= 2
f+(
1) = lim
x3
+
(1 2x) 1x
+ 1= lim
x3+
2(x+ 1)x+ 1
= 2
f
(1) =f
+(1) = 2f(x) is differentiable at x= 1 and f(1) = 2.
Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 34 / 34
http://find/