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Printer QR in Software Calculate the following derivatives: d dx Sin 1 x

Calculate the following derivatives: d dx Sin 1 x
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Sin 1 (x 2 + x)
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Sin 1
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SOLUTION We have d Sin 1 x dx d Sin 1 x 2 + x dx d Sin 1 (1/x) dx
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1 = 1 x2 = 1 1 (x 2 1
x= 2/2
2, 15 = , 65
x=1/3
+ x)2
(2x + 1)
x=1/3
x= 3
1 (1/x)2
1 x2
x= 3
1 = 6
CHAPTER 6 Transcendental Functions
You Try It: Calculate (d/dx)Cos 1 [x 2 + x] Also calculate (d/dx)Sin 1 [ln x x 3 ] EXAMPLE 637
Calculate each of the following derivatives: d dx Cos 1 x
x =1/2
d dx
Cos 1 (ln x)
x= e
d dx
Cos 1 ( x)
x =1/2
SOLUTION We have d Cos 1 x dx d Cos 1 (ln x) dx d Cos 1 ( x) dx 1 = 1 x2 = =
x=1/2
x=1/2
x=1/2
2 = , 3 1 x
x= e
1 1 (ln x)2 1
x= e
2 = , 3e = 1
x=1/2
1 ( x)2
1 1/2 x 2
You Try It: Calculate (d/dx) ln[Cos 1 x] and (d/dx) exp[Sin 1 x]
THE INVERSE TANGENT FUNCTION
De ne the function Tan x to be the restriction of tan x to the interval ( /2, /2) Observe that the tangent function is unde ned at the endpoints of this interval Since d Tan x = sec2 x dx we see that Tan x is increasing, hence it is one-to-one (Fig 619) Also Tan takes arbitrarily large positive values when x is near to, but less than, /2 And Tan takes negative values that are arbitrarily large in absolute value when x is near to, but greater than, /2 Therefore Tan takes all real values Since Tan : ( /2, /2) ( , ) is one-to-one and onto, the inverse function Tan 1 : ( , ) ( /2, /2) exists The graph of this inverse function is shown in Fig 620 It is obtained by the usual procedure of re ecting in the line y = x EXAMPLE 638
Calculate Tan 1 1, Tan 1 1/ 3,
Tan 1 ( 3)
Transcendental Functions
y = Tan x
Fig 619
y = Tan
Fig 620
SOLUTION We have Tan 1 1 = , 4 1 Tan 1/ 3 = , 6 1 Tan ( 3) = 3
CHAPTER 6 Transcendental Functions
As with the rst two trigonometric functions, we note that the tangent func tion takes each of the values 1, 1/ 3, 3 at many different points of its domain But Tan x takes each of these values at just one point of its domain The derivative of our new function may be calculated in the usual way The result is d 1 Tan 1 t = dt 1 + t2 Next we calculate some derivatives: EXAMPLE 639
Calculate the following derivatives: d dx Tan 1 x
x =1
d dx
Tan 1 (x 3 )
x= 2
d dx
Tan 1 (ex )
x =0
SOLUTION We have d Tan 1 x dx d Tan 1 (x 3 ) dx =
1 1 + x2
1 = , 2
x= 2
x= 2
1 3x 2 3 )2 1 + (x 1 ex 1 + (ex )2
2 = , 3
d Tan 1 (ex ) dx
1 = 2
You Try It: Calculate (d/dx)Tan 1 [ln x + x 3 ] and (d/dx) ln[Tan 1 x]
INTEGRALS IN WHICH INVERSE TRIGONOMETRIC FUNCTIONS ARISE
Our differentiation formulas for inverse trigonometric functions can be written in reverse, as antidifferentiation formulas We have du = Sin 1 u + C; 1 u2 du = Cos 1 u + C; 2 1 u du du = Tan 1 u + C 1 + u2 The important lesson here is that, while the integrands involve only polynomials and roots, the antiderivatives involve inverse trigonometric functions
EXAMPLE 640
Transcendental Functions
Evaluate the integral sin x 1 + cos2 x dx
SOLUTION For clarity we set (x) = cos x, (x) = sin x The integral becomes (x) dx 1 + 2 (x)
By what we have just learned about Tan 1 , this last integral is equal to Tan 1 (x) + C Resubstituting (x) = cos x yields that sin x dx = Tan 1 (cos x) + C 1 + cos2 x x/(1 + x 4 ) dx
You Try It: Calculate EXAMPLE 641
Calculate the integral
SOLUTION For clarity we set (x) = x 3 , (x) = 3x 2 The integral then becomes (x) dx 1 2 (x) We know that this last integral equals Sin 1 (x) + C Resubstituting the formula for gives a nal answer of 3x 2 dx = Sin 1 (x 3 ) + C 6 1 x You Try It: Evaluate the integral x dx 1 x4
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