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240_James Stewart Calculus 5 Edition Answers

240_James Stewart Calculus 5 Edition Answers - Stewart...

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Unformatted text preview: Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.4 Derivatives of Trigonometric Functions lim x 1 sin (x 1) 2 sin (x 1) 1 sin (x 1) 1 1 = lim lim = 1= x1 3 3 1 ( x+2)( x 1) x 1 x+2 x 1 = lim x x +x 2 d d sin x 45. (a) tan x= dx dx cos x d d 1 sec x= (b) dx dx cos x (c) 2 sec x= sec xtan x= cos xcos x sin x ( sin x) 2 cos x (cos x)(0) 1( sin x) 2 2 = 2 cos x+sin x 2 cos x . So sec xtan x= cos x 1 2 . So sec x= 2 . cos x sin x 2 . cos x d d 1+cot x (sin x+cos x)= dx dx x 2 cos x sin x = x( csc x) (1+cot x)( xcot x) 2 2 = x [ csc x+(1+cot x) cot x] 2 csc x 2 csc x 2 csc x+cot x+cot x 1+cot x = = x x So cos x sin x= cot x 1 . x 46. Let PR =x . Then we get the following formulas for r and h in terms of and x : sin 2 = r x h 12 1 r =xsin and cos = h=xcos . Now A ( ) = r and B ( ) = (2r )h=rh . So 2 2x 2 2 2 12 A( ) r 2 1 r1 xsin ( /2 ) lim = lim = lim + B ( ) = lim rh 2 2 0 + +h + xcos ( /2 ) 0 = 1 2 0 0 lim tan ( /2 ) =0 . + 0 47. By the definition of radian measure, s=r , where r is the radius of the circle. 8 ...
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