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Done by , ASHAMS KURIAN 11 B K.V.PATTOM
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Done BY, Achuthan xi b k.v.pattom
Limits and Derivatives
Concept of a Function
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
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.
Notation for a Function : f(x)
The Idea of Limits
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)
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
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
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
Consider the function
The Idea of Limits
x
xxh )(
x -4 -3 -2 -1 0 1 2 3 4
g(x)
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
1)(lim0
xhx
1)(lim0
xhx
)(lim0
xhx does not
exist.
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
Theorems On Limits
Theorems On Limits
Theorems On Limits
Theorems On Limits
Limits at Infinity
Limits at Infinity
Consider1
1)(
2
xxf
Generalized, if
)(lim xfx
then
0)(
lim xf
kx
Theorems of Limits at Infinity
Theorems of Limits at Infinity
Theorems of Limits at Infinity
Theorems of Limits at Infinity
Theorem
where θ is measured in radians.
All angles in calculus are measured in radians.
1sin
lim0
The Slope of the Tangent to a Curve
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
Increments
The increment △x of a variable is the change in x from a fixed value x = x0 to another value x = x1.
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).
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
The derivative of a function y = f(x) with respect to x is usually denoted by
,dx
dy),(xf
dx
d ,'y ).(' xf
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.
The value of the derivative of y = f(x) with respect to x at x = x0 is denoted
by or .0xxdx
dy
)(' 0xf
To obtain the derivative of a function by its definition is called differentiation of the function from first principles.
• 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
• 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
• 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
• 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
• 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
• 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
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
• 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
• 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
• Using the same methods as in
the proof of Formula 4, we can prove:
(cos ) sind
x xdx
Formula 5DERIV. OF COSINE FUNCTION
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
• 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
• 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
• 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
• 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
– 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:
• 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
• 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:
• 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
THANK YOU