Limits And Derivative
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Transcript of Limits And Derivative
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Done BY, Achuthan xi b k.v.pattom
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Limits and Derivatives
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Concept of a Function
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y is a function of x, and the relation y = x2 describes a function. We notice that with such a relation, every value of x corresponds to one (and only one) value of y.
y = x2
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Since the value of y depends on a given value of x, we call y the dependent variable and x the independent variable and of the function y = x2.
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Notation for a Function : f(x)
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The Idea of Limits
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Consider the function
The Idea of Limits
2
4)(
2
x
xxf
x 1.9 1.99 1.999 1.9999 2 2.0001 2.001 2.01 2.1
f(x)
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Consider the function
The Idea of Limits
2
4)(
2
x
xxf
x 1.9 1.99 1.999 1.9999 2 2.0001 2.001 2.01 2.1
f(x) 3.9 3.99 3.999 3.9999 un-defined
4.0001 4.001 4.01 4.1
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Consider the function
The Idea of Limits 2)( xxg
x 1.9 1.99 1.999 1.9999 2 2.0001 2.001 2.01 2.1
g(x) 3.9 3.99 3.999 3.9999 4 4.0001 4.001 4.01 4.1
2)( xxg
x
y
O
2
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If a function f(x) is a continuous at x0,
then . )()(lim 00
xfxfxx
4)(lim2
xfx
4)(lim2
xgx
approaches to, but not equal to
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Consider the function
The Idea of Limits
x
xxh )(
x -4 -3 -2 -1 0 1 2 3 4
g(x)
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Consider the function
The Idea of Limits
x
xxh )(
x -4 -3 -2 -1 0 1 2 3 4
h(x) -1 -1 -1 -1 un-defined
1 2 3 4
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1)(lim0
xhx
1)(lim0
xhx
)(lim0
xhx does not
exist.
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A function f(x) has limit l at x0 if f(x) can be made as close to l as we please by taking x sufficiently close to (but not equal to) x0. We write
lxfxx
)(lim0
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Theorems On Limits
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Theorems On Limits
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Theorems On Limits
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Theorems On Limits
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Limits at Infinity
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Limits at Infinity
Consider1
1)(
2
xxf
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Generalized, if
)(lim xfx
then
0)(
lim xf
kx
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Theorems of Limits at Infinity
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Theorems of Limits at Infinity
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Theorems of Limits at Infinity
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Theorems of Limits at Infinity
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Theorem
where θ is measured in radians.
All angles in calculus are measured in radians.
1sin
lim0
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The Slope of the Tangent to a Curve
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The Slope of the Tangent to a Curve
The slope of the tangent to a curve y = f(x) with respect to x is defined as
provided that the limit exists.
x
xfxxf
x
yAT
xx
)()(limlim of Slope
00
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Increments
The increment △x of a variable is the change in x from a fixed value x = x0 to another value x = x1.
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For any function y = f(x), if the variable x is given an increment △x from x = x0, then the value of y would change to f(x0 + △x) accordingly. Hence thee is a corresponding increment of y(△y) such that △y = f(x0 + △x) –
f(x0).
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Derivatives(A) Definition of Derivative.
The derivative of a function y = f(x) with respect to x is defined as
provided that the limit exists.
x
xfxxf
x
yxx
)()(limlim
00
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The derivative of a function y = f(x) with respect to x is usually denoted by
,dx
dy),(xf
dx
d ,'y ).(' xf
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The process of finding the derivative of a function is called differentiation. A function y = f(x) is said to be differentiable with respect to x at x = x0 if the derivative of the function with respect to x exists at x = x0.
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The value of the derivative of y = f(x) with respect to x at x = x0 is denoted
by or .0xxdx
dy
)(' 0xf
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To obtain the derivative of a function by its definition is called differentiation of the function from first principles.
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• Let’s sketch the graph of the function f(x) = sin
x, it looks as if the graph of f’ may be the same
as the cosine curve.
DERIVATIVES OF TRIGONOMETRIC FUNCTIONS
Figure 3.4.1, p. 149
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• From the definition of a derivative, we have:
0 0
0
0
0
0 0 0
( ) ( ) sin( ) sin'( ) lim lim
sin cos cos sin h sinlim
sin cos sin cos sinlim
cos 1 sinlim sin cos
cos 1limsin lim lim cos lim
h h
h
h
h
h h h h
f x h f x x h xf x
h hx h x x
hx h x x h
h h
h hx x
h h
hx x
h
0
sin h
h
DERIVS. OF TRIG. FUNCTIONS Equation 1
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• Two of these four limits are easy to evaluate.
DERIVS. OF TRIG. FUNCTIONS
0 0 0 0
cos 1 sinlimsin lim lim cos limh h h h
h hx x
h h
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• Since we regard x as a constant
when computing a limit as h → 0,
we have:
DERIVS. OF TRIG. FUNCTIONS
limh 0
sin x sin x
limh 0
cos x cos x
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• The limit of (sin h)/h is not so obvious.
• In Example 3 in Section 2.2, we made
the guess—on the basis of numerical and
graphical evidence—that:
0
sinlim 1
DERIVS. OF TRIG. FUNCTIONS Equation 2
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• We can deduce the value of the remaining
limit in Equation 1 as follows.
0
0
2
0
cos 1lim
cos 1 cos 1lim
cos 1
cos 1lim
(cos 1)
DERIVS. OF TRIG. FUNCTIONS
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2
0
0
0 0
0
sinlim
(cos 1)
sin sinlim
cos 1
sin sin 0lim lim 1 0
cos 1 1 1
cos 1lim 0
DERIVS. OF TRIG. FUNCTIONS Equation 3
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• If we put the limits (2) and (3) in (1),
we get:
• So, we have proved the formula for sine,
0 0 0 0
cos 1 sin'( ) limsin lim lim cos lim
(sin ) 0 (cos ) 1
cos
h h h h
h hf x x x
h hx x
x
DERIVS. OF TRIG. FUNCTIONS Formula 4
(sin ) cosd
x xdx
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• Differentiate y = x2 sin x.– Using the Product Rule and Formula 4,
we have:
2 2
2
(sin ) sin ( )
cos 2 sin
dy d dx x x x
dx dx dx
x x x x
Example 1DERIVS. OF TRIG. FUNCTIONS
Figure 3.4.3, p. 151
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• Using the same methods as in
the proof of Formula 4, we can prove:
(cos ) sind
x xdx
Formula 5DERIV. OF COSINE FUNCTION
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2
2
2 22
2 2
2
sin(tan )
cos
cos (sin ) sin (cos )
coscos cos sin ( sin )
cos
cos sin 1sec
cos cos
(tan ) sec
d d xx
dx dx x
d dx x x x
dx dxx
x x x x
x
x xx
x xd
x xdx
DERIV. OF TANGENT FUNCTION Formula 6
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• We have collected all the differentiation
formulas for trigonometric functions here. – Remember, they are valid only when x is measured
in radians.
2 2
(sin ) cos (csc ) csc cot
(cos ) sin (sec ) sec tan
(tan ) sec (cot ) csc
d dx x x x x
dx dxd d
x x x x xdx dxd d
x x x xdx dx
DERIVS. OF TRIG. FUNCTIONS
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• Differentiate
• For what values of x does the graph of f have
a horizontal tangent?
sec( )
1 tan
xf x
x
Example 2DERIVS. OF TRIG. FUNCTIONS
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• The Quotient Rule gives:
2
2
2
2 2
2
2
(1 tan ) (sec ) sec (1 tan )'( )
(1 tan )
(1 tan )sec tan sec sec
(1 tan )
sec (tan tan sec )
(1 tan )
sec (tan 1)
(1 tan )
d dx x x x
dx dxf xx
x x x x x
x
x x x x
x
x x
x
Example 2Solution:
tan2 x + 1 = sec2 x
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• Find the 27th derivative of cos x.
– The first few derivatives of f(x) = cos x are as follows:
(4)
(5)
'( ) sin
''( ) cos
'''( ) sin
( ) cos
( ) sin
f x x
f x x
f x x
f x x
f x x
Example 4DERIVS. OF TRIG. FUNCTIONS
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– We see that the successive derivatives occur in a cycle of length 4 and, in particular, f (n)(x) = cos x whenever n is a multiple of 4.
– Therefore, f (24)(x) = cos x
– Differentiating three more times, we have:
f (27)(x) = sin x
Example 4Solution:
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• Find
– In order to apply Equation 2, we first rewrite the function by multiplying and dividing by 7:
0
sin 7lim
4x
x
x
sin 7 7 sin 7
4 4 7
x x
x x
Example 5DERIVS. OF TRIG. FUNCTIONS
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• If we let θ = 7x, then θ → 0 as x → 0.
So, by Equation 2, we have:
0 0
0
sin 7 7 sin 7lim lim
4 4 7
7 sinlim
4
7 71
4 4
x x
x x
x x
Example 5Solution:
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• Calculate .
– We divide the numerator and denominator by x:
by the continuity of cosine and Eqn. 2
0lim cotx
x x
Example 6DERIVS. OF TRIG. FUNCTIONS
0 0 0
0
0
cos coslim cot lim lim
sinsin
lim cos cos0sin 1lim
1
x x x
x
x
x x xx x
xxx
x
x
x
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