R7 a i:*=$I #ii. f(x) = S(h(x)) =+ f'(x) = g'(h(x)) . h'(x)iii. The derivative of a composite function is the

derivative of the outside function with respect tothe inside function, times the derivative of theinside function with respect to x.

b. See the derivation in the text. This derivationconstitutes a proof.Au must be nonzero throughout the interval.

c. i. l(x) = (x2 - 4\3f '(x) = 31xz - 4)2 . 2x = 6x(x2 - 4)2

ii. f(x) = xG - 12xa + 48x2 -64f'(x)=6xs-48x3+96xExpanding the answer to part (i) givesf'(x) = gxs - 48x3 + 96x, which checks.

d. i. f'(x) = -3x2 sin x3ii.g'(x)=5cos-Sxiii. fr'(x) = 6 coss x (-sin x) = -6 sin x coss xiv. k'(x) = 0

e. Shark ProblemW = 0.6x3 and dx/dt = 0.4dW dff dxff = ff . ff = 1.8x2 . 0.4 = 0.72x2

It x=2, W = 0.6.23 = 4.8 lbdWdt= 0.72(21=2.88Shark is gaining about 2.88 lb/day.lf x = 10, W = 0.6.103 = 600 lb.dWdt= O1zfiO\=72Shark is gaining aboulT2lb/day.The chain rule is used to get dWdt from dWdx bymultiplying the latter by dx/dt.

R8. a. lim -U=1.x+0 x

b. x (sin x)/x-0.05 0.99958338541...-0.04 0.99973335466...-0.03 0.99985000674...-0.02 0.99993333466...-0.01 0.99998333341...0.00 (No value)0.01 0.99998333341...a.02 0.99993333466...0.03 0.99985000674...0.04 0.99973335466...0.05 0.99958338541...

The values approach 1 as x approaches 0.c. See text statement of squeeze theorem.

Squeeze (sin x)/x between cos x and sec x.d. See proof in text (Section 3-8).e. cos x = sin (r/2 - x)

cos'x = cos (n/2 - x) (-1) = -sin x, Q.E.D.

f. Clock Problemd(t)=C+AcosB(t-D)C= 180, A=20D = 0 for cosine since hand starts at a high point.B = 2trl6O = rcl30 since period is 60 seconds.

d(t) = 180 + 20 cos ft to'1t; = -ffsin fttAt 2, t = 10. d'(10) = -1.81 cm/secAt 3, t = 15. d'(15) = -2.O9 cm/secAt 7, t = 35. d'(35) = 1.05 cm/secAt the 2 and 3, the tip is going down, so the distancefrom the floor is decreasing, which is implied by thenegative derivatives.At the 7, the tip is going up, as implied by thepositive derivative.

R9. a. f(x) = 36x5 + f(x) = 6x616b. dyldx = sin 0.2x and y(0) = 3

Y=-5cos0.2x+C3=-5cos0+C + C=8i.Y=5cos0.2x+8

c. Distance Problem

Y(t)=4tY2+C100=4(032)+C + C=1oo;. Y(t) = 413/2 * 1ggy(60) = 4(AOTsrz + 100 = 1959.03...v(60) = 6(0Otrz1 = 46.475.'She is about 1959 feet away from the house, goingabout 46.5 ftlsec 1 minute after she starts.

d. An antiderivative and an indefinite integral are thesam'e tning.

Concepts Problems

C1. lntroduction to the Derivative of a Producta. f(x) = x7, g(x) = xe. So h(x) = f(x).g(x) = x16.

b' h'(x) = 16x15.c. f'(x) = 7x6 , g'(x) = 918.

So f'(x)'g'(x) = 63x14 * h'(x).d. h'(x) = f'(x).9(x) + f(x)'g'(x)

=/v6.19 + x7.gx8 = 16x15, which checks.

C2. Graoh of an lnterestino Functionx-sin2x

a. tfxl = ---------;-SIN X

f(0) has the form 0/0, which is indeterminate.f is discontinuous at x = 0 because f(0) does notexist.

b. By graph (below) or by reele, f(x) seems toapproach -1 as x approaches zero.Define f(0) to be -1.

Conjecture: The function ls differentiable at x = 0.The derivative should equal zero because the graphis horizontal at x = 0.

Problem Set 3-10 Solutions Monuol 43