# F x sin x f x cos x f x sin x and f 4 x

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Unformatted text preview: ) = −x2 + 4 if |x| &lt; 2 so f is diﬀerentiable everywhere except perhaps at ±2. f is continuous at −2 and 2, also lim− f (x) = lim− (−2x) = −4 and lim+ f (x) = x→2 x→2 x→2 lim+ (2x) = 4 so f is not diﬀerentiable at x = 2. Similarly, f is not diﬀerentiable at x = −2. x→2 80. f (x) = −(1)x−2 , f (x) = (2 · 1)x−3 , f (x) = −(3 · 2 · 1)x−4 n(n − 1)(n − 2) · · · 1 f (n) (x) = (−1)n xn+1 −3 (b) f (x) = −2x , f (x) = (3 · 2)x−4 , f (x) = −(4 · 3 · 2)x−5 (n + 1)(n)(n − 1) · · · 2 f (n) (x) = (−1)n xn+2 81. (a) (a) d d d2 d d d d d2 [cf (x)] = [cf (x)] = c [f (x)] = c [f (x)] = c 2 [f (x)] dx2 dx dx dx dx dx dx dx d d d2 d d2 d2 d d [f (x) + g (x)] = [f (x)] + [g (x)] = 2 [f (x)] + 2 [g (x)] [f (x) + g (x)] = dx2 dx dx dx dx dx dx dx (b) yes, by repeated application of the procedure illustrated in part (a) 82. (f · g ) = f g + gf , (f · g ) = f g + g f + gf + f g = f g + 2f g + f g 83. (a) f (x) = nxn−1 , f (x) = n(n − 1)xn−2 , f (x) = n(n − 1)(n − 2)xn−3 , . . ., f (n) (x) = n(n − 1)(n − 2) · · · 1 (b) from part (a), f (k) (x) = k (k − 1)(k − 2) · · · 1 so f (k+1) (x) = 0 thus f (n) (x) = 0 if n &gt; k 79 Chapter 3 (c) from parts (a) and (b), f (n) (x) = an n(n − 1)(n − 2) · · · 1 84. f (2 + h) − f (2) = f (2); f (x) = 8x7 − 2, f (x) = 56x6 , so f (2) = 56(26 ) = 3584. h→0 h 85. (a) lim If a function is diﬀerentiable at a point then it is continuous at that point, thus f is continuous on (a, b) and consequently so is f . (b) f and all its derivatives up to f (n−1) (x) are continuous on (a, b) EXERCISE SET 3.4 1. f (x) = −2 sin x − 3 cos x 2. f (x) = sin x(− sin x) + cos x(cos x) = cos2 x − sin2 x = cos 2x 3. f (x) = 4. f (x) = x2 (− sin x) + (cos x)(2x) = −x2 sin x + 2x cos x x(cos x) − sin x(1) x cos x − sin x = x2 x2 5. f (x) = x3 (cos x) + (sin x)(3x2 ) − 5(− sin x) = x3 cos x + (3x2 + 5) sin x 7. cos x x(− csc2 x) − (cot x)(1) x csc2 x + cot x cot x (because = cot x), f (x) = =− x sin x x2 x2 √ f (x) = sec x tan x − 2 sec2 x 8. f (x) = (x2 + 1) sec x tan x + (sec x)(2x) = (x2 + 1) sec x tan x + 2x sec x 9. f (x) = sec x(sec2 x) + (tan x)(sec x tan x) = sec3 x + sec x tan2 x 6. f (x) = 10. f (x) = = 11. (1 + tan x)(sec x tan x) − (sec x)(sec2 x) sec x tan x + sec x tan2 x − sec3 x = (1 + tan x)2 (1 + tan x)2 sec x(tan x + tan2 x − sec2 x) sec x(tan x − 1) = 2 (1 + tan x) (1 + tan x)2 f (x) = (csc x)(− csc2 x) + (cot x)(− csc x cot x) = − csc3 x − csc x cot2 x 12. f (x) = 1 + 4 csc x cot x − 2 csc2 x 13. f (x) = (1 + csc x)(− csc2 x) − cot x(0 − csc x cot x) csc x(− csc x − csc2 x + cot2 x) = but 2 (1 + csc x) (1 + csc x)2 1 + cot2 x = csc2 x (identity) thus cot2 x − csc2 x = −1 so csc x(− csc x − 1) csc x =− f (x) = (1 + csc x)2 1 + csc x csc x(1 + sec2 x) tan x(− csc x cot x) − csc x(sec2 x) =− 2x tan tan2 x 14. f (x) = 15. f (x) = sin2 x + cos2 x = 1 (identity) so f (x) = 0 16. f (x) = 1 = tan x, so f (x) = sec2 x cot x Exercise Set 3.4 17. tan x (because sin x sec x = (sin x)(1/ cos x) = tan x), 1 + x tan x (1 + x tan x)(sec2 x) − tan x[x(sec2 x) + (tan x)(1)] f (x) = (1 + x tan x)2 sec2 x − tan2 x 1 = = (because sec2 x − tan2 x = 1) (1 + x tan x)2 (1 + x tan x)2 f (x) = f (x) = (x2 + 1) cot x (because cos x csc x = (cos x)(1/ sin x) = cot x), 3 − cot x f (x) = 18. 80 (3 − cot x)[2x cot x − (x2 + 1) csc2 x] − (x2 + 1) cot x csc2 x (3 − cot x)2 = 6x cot x − 2x cot2 x − 3(x2 + 1) csc2 x (3 − cot x)2 19. dy/dx = −x sin x + cos x, d2 y/dx2 = −x cos x − sin x − sin x = −x cos x − 2 sin x 20. dy/dx = − csc x cot x, d2 y/dx2 = −[(csc x)(− csc2 x) + (cot x)(− csc x cot x)] = csc3 x + csc x cot2 x 21. dy/dx = x(cos x) + (sin x)(1) − 3(− sin x) = x cos x + 4 sin x, d2 y/dx2 = x(− sin x) + (cos x)(1) + 4 cos x = −x sin x + 5 cos x 22. dy/dx = x2 (− sin x) + (cos x)(2x) + 4 cos x = −x2 sin x + 2x cos x + 4 cos x, d2 y/dx2 = −[x2 (cos x) + (sin x)(2x)] + 2[x(− sin x) + cos x] − 4 sin x = (2 − x2 ) cos x − 4(x + 1) sin x 23. dy/dx = (sin x)(− sin x) + (cos x)(cos x) = cos2 x − sin2 x, d2 y/dx2 = (cos x)(− sin x) + (cos x)(− sin x) − [(sin x)(cos x) + (sin x)(cos x)] = −4 sin x cos x 24. dy/dx = sec2 x; d2 y/dx2 = 2 sec2 x tan x 26. Let f (x) = sin x, then f (x) = cos x. (a) f (0) = 0 and f (0) = 1 so y − 0 = (1)(x − 0), y = x (b) (c) 27. f (π ) = 0 and f (π ) = −1 so y − 0 = (−1)(x − π ), y = −x + π 1 1 1 π 1 1 π 1 π π = √ and f = √ so y − √ = √ x − , y = √ x− √ + √ f 4 4 4 2 2 2 2 2 42 2 Let f (x) = tan x, then f (x) = sec2 x. f (0) = 0 and f (0) = 1 so y − 0 = (1)(x − 0), y = x. π π π = 1 and f = 2 so y − 1 = 2 x − , y = 2x − + 1. (b) f 4 4 4 2 π π π π = −1 and f − = 2 so y + 1 = 2 x + , y = 2x + − 1. (c) f − 4 4 4 2 (a) 29. (a) If y = cos x then y = − sin x and y = − cos x so y + y = (− cos x) + (cos x) = 0; if y = sin x then y = cos x and y = − sin x so y + y = (− sin...
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