dy dx = 1 2 1 q x +p x

Homework Problem Solution

§3.4 42.

dy dx = 1

2

1 q

x +p x +√

x d dx(x +

q x +√

x)

= 1

2 q

x +p x +√

x

[1 + 1

2p x +√

x d

dx(x +√ x)]

= 1

2 q

x +p x +√

x

[1 + 1

2p x +√

x(1 + 1 2√

x)]

60

dy

dx = 2 cos 2x − 2 cos x = 2 cos²x − 1 − cos x = 0 when cos x = ¹₂ or cos x = 1

⇒ x = 2nπ or x = (2n + 1)π ± ^π₃

Thus the tangent lines are horizontal at these points.

63.

(a) h⁰(x) = f⁰(g(x))g⁰(x) ⇒ h⁰(1) = f⁰(g(1))g⁰(1) = f⁰(2)g⁰(1) = 30.

(b) H⁰(x) = g⁰(f (x))f⁰(x) ⇒ H⁰(1) = g⁰(f (1))f⁰(1) = g⁰(3)f⁰(1) = 36.

65.

(a) u⁰(x) = f⁰(g(x))g⁰(x) ⇒ u⁰(1) = f⁰(g(1))g⁰(1) = f⁰(3)g⁰(1) = (−¹₄)(−3) =

3 4.

(b) v⁰(x) = g⁰(f (x))f⁰(x) ⇒ v⁰(1) = g⁰(f (1))f⁰(1) = g⁰(2)f⁰(1) = 36 which does not exist since g⁰(2) does not exist.

(c) w⁰(x) = g⁰(g(x))g⁰(x) ⇒ w⁰(1) = g⁰(g(1))g⁰(1) = g⁰(3)g⁰(1) = ²₃(−3) =

−2.

The above derivative values are obtained by observing the slopes.

67.

From the figure we know that f (3) = 2 and f⁰(3) = −²₃. g⁰(x) = ¹

2√

f (x)f⁰(x) ⇒ g⁰(3) = ¹

2√

2(−²₃) = − ¹

3√ 2

1

74.

F⁰(x) = f⁰(x · f (xf (x))) · d

dx(x · f (xf (x)))

= f⁰(x · f (xf (x)))[1 · f (xf (x)) + x · f⁰(xf (x)) · (f (x) + xf⁰(x))]

F⁰(1) = f⁰(1 · f (1 · 2))[f (1 · 2) + 1 · f⁰(1 · 2) · (2 + 1 · 4)]

= f⁰(3)[3 + 5 · 6]

= 198

91.

(a) f is even ⇔ f (−x) = f (x). Differentiating both sides using the Chain Rule, we have −f⁰(−x) = f⁰(x) ⇒ f⁰(−x) = −f⁰(x), which means f⁰(x) is an odd function.

(b) f is odd ⇔ f (−x) = −f (x). Differentiating both sides using the Chain Rule, we have −f⁰(−x) = −f⁰(x) ⇒ f⁰(−x) = f⁰(x), which means f⁰(x) is an even function.

93. (a) d

dx(sinⁿx cos(nx)) = n sinⁿ⁻¹x cos x cos(nx) − sinⁿx · n sin(nx)

= n sinⁿ⁻¹x[cos x cos(nx) − sin x sin(nx)]

= n sinⁿ⁻¹x cos[(n + 1)x]

(b) d

dx(cosⁿx cos(nx)) = −n cosⁿ⁻¹x sin x cos(nx) − cosⁿx · n sin(nx)

= −n cosⁿ⁻¹x[sin x cos(nx) + cos x sin(nx)]

= −n cosⁿ⁻¹x sin[(n + 1)x]

96.

(a) _dx^d(x²)¹² = ¹

2(x²)¹² · 2x = _|x|^x for x 6= 0. |x| is not differentiable at x = 0.

(b) f (x) = | sin x| ⇒ f⁰(x) = _{| sin x|}^{sin x} cos x by the Chain Rule.

f⁰(x) =

 cos x if sin x > 0

− cos x if sin x < 0

f is not differentiable when x = nπ, n ∈ Z.

2

(c) g(x) = sin(|x|) ⇒ g⁰(x) = cos(|x|)_|x|^x by the Chain Rule.

g⁰(x) =

 cos x if x > 0

− cos x if x < 0 g is not differentiable at x = 0.

98.

dy dx = dy

du du dx d²y

dx² = (d²y du²

du dx)du

dx + dy du

d²u dx²

= d²y du²(du

dx)²+ dy du

d²u dx² d³y

dx³ = [d³y du³(du

dx)](du

dx)²+ d²y du² · 2du

dx d²u

dx² + d²y du²

du dx

d²u dx² + dy

du d³u dx³

= d³y du³(du

dx)³+ 3d²y du²

d²u dx²

du dx + dy

du d³u dx³

47.

lim

x→^π₄

1 − tan x

sin x − cos x = lim

x→^π₄

cos x − sin x

(sin x − cos x)(cos x) = −√ 2

48.

x→1lim

sin(x − 1)

x²+ x − 2 = lim

x→1

sin(x − 1)

(x − 1)(x + 2) = (lim

x→1

sin (x − 1) x − 1 )(lim

x→1

1

x + 2) = 1 3 3

since both limits exist.

50.

Let h(x) = sin x and f (x) = xh(x), we have f⁰(x) = h(x) + xh⁰(x) f⁰⁰(x) = 2h⁰(x) + xh⁰⁰(x) f⁰⁰(x) = 3h⁰⁰(x) + xh⁰⁰⁰(x)

...

f⁽ⁿ⁾(x) = nh⁽ⁿ⁻¹⁾(x) + xh⁽ⁿ⁾(x)

Thus,

d³⁵

dx³⁵(x sin x) = 35 d³⁴

dx³⁴(sin x) + x d³⁵

dx³⁵(sin x)

= −35 sin x − x cos x

54.

The radius of the semicircle is r = 10 sin^θ₂.

A(θ) = 1

2πr² = 50π sinθ 2 B(θ) = 2 · 1

2(10 sinθ

2)(10 cosθ 2)

= 100 sinθ 2cosθ

2 lim

θ→0⁺

A(θ)

B(θ) = lim

θ→0⁺

π sin^θ₂ 2 cos ^θ₂

= 0

4