# barcode in ssrs 2008 Thus, f (x) = 5x 3 2 has an inverse, f 1 (y) = and so, by Problem 37.1, f is one-one. in .NET framework Encode Quick Response Code in .NET framework Thus, f (x) = 5x 3 2 has an inverse, f 1 (y) = and so, by Problem 37.1, f is one-one.

Thus, f (x) = 5x 3 2 has an inverse, f 1 (y) = and so, by Problem 37.1, f is one-one.
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37.3 Find formulas for the inverses of the following one-one functions: (a) f (x) = 10x 4 (b) f (x) = 3ex + 1
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y+4 =x 10 y+4 f 1 (y) = 10 or
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(a) Let y = 10x 4. Then, y + 4 = 10x Hence, (b) Let y = 3ex + 1. Then, y 1 = 3ex Hence, or y 1 = ex 3 f 1 (y) = ln y 1 3 or ln y 1 =x 3
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37.4 Find the following values: 1 1 (a) sin 1 (b) cos 1 2 2
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(a) sin 1 1 2
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(c) tan 1 ( 1)
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1 is the angle in , whose sine is , that is, . 2 2 2 6
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CHAP. 37]
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INVERSE TRIGONOMETRIC FUNCTIONS
1 2 is the angle in [0, ] whose cosine is , that is, . 2 3 whose tangent is 1, that is, . (c) tan 1 ( 1) = the angle in , 2 2 4 (b) cos 1 1 2
37.5 Evaluate:
(a) cot 1
(b) sec 1 2;
(c) csc 1 2/ 3 .
(a) cot 1 3 is the angle between 0 and such that cot = 3. This angle is /6 (see Fig. 37-8). Note that an inverse trigonometric function must be represented by an angle in radians. (b) The angle whose secant is +2 must lie in the rst quadrant. The secant function is negative in the third quadrant. From Fig. 37-8, = /3. (c) The angle whose cosecant is +2/ 3 must lie in the rst quadrant. The cosecant function is negative in the third quadrant. From Fig. 37-8, = /3.
Fig. 37-8 37.6 Find the derivatives of: (a) sin 1 2x (b) cos 1 x 2
(c) tan 1
(d) sec 1 x 3
(a) Dx (sin 1 2x) = = (b) Dx (cos 1 x 2 ) = = (c) Dx tan 1
Dx (2x) [by the chain rule] 1 (2x)2 1 2 (2) = 1 4x 2 1 4x 2 Dx (x 2 ) [by the chain rule] 1 (x 2 )2 2x 1 (2x) = 4 1 x 1 x4 [by the chain rule] 4 1 4 1 + x2
1 x x = Dx x 2 2 2 1+ 2 = = 1 2 1 + x4 4 4 + x2 1 1 2 = 1 2
2 = 4 + x2
(d) Dx (sec 1 x 3 ) =
Dx (x 3 ) [by the chain rule] (x 3 )2 1 3 1 (3x 2 ) = = 3 x6 1 x x x6 1 x3
INVERSE TRIGONOMETRIC FUNCTIONS
[CHAP. 37
1 . 37.7 Assuming that sec 1 x is differentiable, verify that Dx (sec 1 x) = x x2 1
Let y = sec 1 x. Then, sec y = x Dx (sec y) = Dx (x) dy =1 [by the chain rule] (sec y tan y) dx dy 1 1 = = dx sec y tan y x tan y But the identity 1 + tan2 y = sec2 y gives 1 + tan2 y = x 2 or tan2 y = x 2 1 or tan y = x 2 1
By de nition of sec 1 x, the angle y lies in the rst or third quadrant, where the tangent is positive. Hence, tan y = + x 2 1 and 1 dy = dx x x2 1
37.8 Prove the following formulas for antiderivatives: (a) (b) (c) x dx 1 = tan 1 + C a a a2 + x 2 dx a2 x2 = sin 1 x +C a [a > 0] [a > 0]
x dx 1 = sec 1 + C 2 a2 a a x x
It is common to write
notation
dx f (x)
1 dx f (x)
In each case, we employ the chain rule to show that the derivative of the function on the right is the integrand on the left. (a) Dx x 1 tan 1 a a x 1 1 1 Dx tan 1 = a a a1+ x 2 a 1 1 1 = 2 = 2 a 1 + x2 a + x2 =
x = (b) Dx sin 1 a = = (c) x 1 Dx sec 1 a a
x 2 1 a
1 a 1 a2 [a > 0 implies a = [ u v = uv] 1
x 2 a 1
x 2 1 a
a2 ]
a2 x 2 1 1 x = Dx sec 1 = a a a x = x 1
x2 a2
1 a2
1 x x 2 a2
CHAP. 37]
INVERSE TRIGONOMETRIC FUNCTIONS
37.9 Find the following antiderivatives: dx dx (b) (a) x2 + 4 9 x2
(a) By Problem 37.8(a), with a = 2,
dx x 2 + 4x + 5
2x dx x 2 2x + 7
x dx 1 = tan + C 2 2 x2 + 4 (b) By Problem 37.8(b), with a = 3, dx 9 x2 (c) Complete the square: x 2 + 4x + 5 = (x + 2)2 + 1. Thus, dx x 2 + 4x + 5 Let u = x + 2. Then du = dx, and dx = (x + 2)2 + 1 du = tan 1 u + C = tan 1 (x + 2) + C u2 + 1 = dx (x + 2)2 + 1 = sin 1 x +C 3
(d) Complete the square: x 2 2x + 7 = (x 1)2 + 6. Let u = x 1, du = dx, 2x = 2u + 2 So, 2x dx = x 2 2x + 7 2u du 2du + u2 + 6 u2 + 6 1 u [by Quick Formula II and Problem 37.8(a)]1 = ln (u2 + 6) + 2 tan 1 + C 6 6 x 1 2 = ln (x 2 2x + 7) + tan 1 + C 6 6 2u + 2 du = u2 + 6
37.10 Evaluate sin (2 cos 1 ( 1 )). 3
Let = cos 1 ( 1 ). Then, by Theorems 26.8 and 26.1, 3 sin 2 = 2 sin cos = 2( 1 cos2 )(cos ) By de nition of the function cos 1 , angle is in the second quadrant (its cosine being negative), and so its sine is positive. Therefore, the plus sign must be taken in the above formula. 1 2 1 sin 2 = 2( 1 cos2 )(cos ) = 2 1 3 3 2 2 8 2 8 1 = = 1 = 3 9 3 9 3 9 4 2 2 2 2 = = 3 3 9