3. g(x) = 5in txConjecture: g'(x) = 3 cos 3xGraph confirms conjecture.

h(x) = 5iP xzCbnjecture: h'(x) = !x qss x2

Graph confirms conjecture.

t(x) = 5iP x0'7ionjecture: t'(x) = 0.7x-0'3 cos x0'7

Graph confirms conjecture!

6. f(x) = sin [g(x)]f is a composite function.g is the inside function.sine is the outside function.Differentiate the outside function with respect to theinside function. Then multiply the answer by thederivative of the inside function with respect to x.

7.a. f(x) = sin 3x. lnside: 3x. Outside: sine.b. lr(x) = sin3 x. lnside: sine. Outside: cube.c. g(x) = sin x3. lnside: cube. Outside: sine.d. r(x) = lcos x.

lnside: cosine. Outside: exponential.e. q(x) = 1/(tan x).

lnside: tangent. Outside: reciprocal.f. L(x) = log (sec x).

lnside: secant. Outside: logarithm'

8. Journal ProblemThe journal entry should mention the conjecturesmade, which ones were right, and what blind alleyswere met on the way to the right conjectures.

Q1.Q2.

Q3.

dv dr duai = ai' o*

b. y' = f'(g(x)) . g'(x)c. To differentiate a composite function, differentiate

the outside function with respect to the insidefunction, then multiply by the derivative of the insidefunction with respect to x.

2.1(x\=(xz-1;sa. f'(x) = 3(x2 - 1)2(2x\b. (i2'- 1)3= xo 3xl + 3x2- 1,

io f'(xi = 6xs -12x3 + 6x.c. From iart a., f'(x) = $x(x+ -2x2 + 1)

= 6x5 -1 2x3 + 6x,so the two answers are equivalent.

3. f(x) = cos 3x =+ l'(x) = -sin 3x . 3

= -3 sin 3x

4. f(x) = sin 5x + f'(x) = 5 cos 5x

5. g(x) = cos (x3) = g'(x) = -3x2 sin (x3)

6. h(x) = sin (x5) + h'(x) = 5x4 cos (x5)

7. y = (cos I)3 + y' = 3(cos x)2 ' (-sin x)

= -3 cosz x sin x

8. f(x) = (sin x)s + f'(x) = S(sin x)a ' cos x

=5sin4xcosx9. y=sin6x + y' = 6 sins xcos x

Q4. No (not continuous)Q6. f(x) = -10x-3Q8. GraPh. (Y = sin x)

Q5. dy/dx - 16x-1/sQ7. Antiderivative

Q9. Graph. (y = cos x)

Q10. Graph. (y - -sin x)

1.a. Let Y = f(u), u = g(x).

36 Colculus: Concepls ond Applicotions Problem Set 3'7