Vv156 Honors Calculus II

Assignment 3

This assignment has a total of (32 points).

Note: Unless specified otherwise, you must show the details of your work via logical reasoning for each exercise.

Simply writing a final result (whether correct or not) will receive 0 point.

Exercise 3.1 [Ste10, p. 205]

(i) (1 pt) The curve y = |x|/

√

2 − x

2 is called a bullet-nose curve. Find an equation of the tangent line to this

curve at the point (1, 1).

(ii) (1 pt) Illustrate part (i) by sketch the curve and the tangent line on the same coordinate system.

(2 pts)

Exercise 3.2 [Ste10, p. 208] Use the Chain Rule to prove the following.

(i) (1 pt) The derivative of an even function is an odd function.

(ii) (1 pt) The derivative of an odd function is an even function.

(2 pts)

Exercise 3.3 [Ste10, p. 208] If y = f(u) and u = g(x), where f and g are twice differentiable functions, show that

d

2y

dx

2

=

d

2y

du

2

du

dx

2

+

dy

du

d

2u

dx

2

(2 pts)

Exercise 3.4 [Ste10, p. 215] Use implicit differentiation to find an equation of the tangent line to the curve at the

given point.

(i) (1 pt) (cardioid) x

2 + y

2 = (2x

2 + 2y

2 − x)

2 at (0,

1

2

).

(ii) (1 pt) (astroid) x

2/3 + y

2/3 = 4 at (−3

√

3, 1).

(iii) (1 pt) (lemniscate) 2(x

2 + y

2

)

2 = 25(x

2 − y

2

) at (3, 1).

(iv) (1 pt) (devil’s curve) y

2

(y

2 − 4) = x

2

(x

2 − 5) at (0, −2).

(4 pts)

Exercise 3.5 [Ste10, Sec. 3.11] Given the following hyperbolic functions defined as

sinh x =

e

x − e

−x

2

cosh x =

e

x + e

−x

2

tanh x =

sinh x

cosh x

csch x =

1

sinh x

sech x =

1

cosh x

coth x =

cosh x

sinh x

(i) (3 pts) Show that

d

dx

(sinh x) = cosh x

d

dx

(cosh x) = sinh x

d

dx

(tanh x) = sech2

x

d

dx

(csch x) = − csch x coth x

d

dx

(sech x) = − sech x tanh x

d

dx

(coth x) = − csch2

x

(ii) (3 pts) and show that1

d

dx

(sinh−1

x) = 1

√

1 + x

2

d

dx

(cosh−1

x) = 1

√

x

2 − 1

d

dx

(tanh−1

x) = 1

1 − x

2

d

dx

(csch−1

x) = −

1

|x|

√

x

2 + 1

d

dx

(sech−1

x) = −

1

x

√

1 − x

2

d

dx

(coth−1

x) = 1

1 − x

2

1Notice that the formulas for the derivatives of tanh−1 x and coth−1 x appear to be identical. But the domains of these functions have

no numbers in common: tanh−1 x is defined for |x| < 1, whereas coth−1 x is defined for |x| > 1.

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(6 pts)

Exercise 3.6 [Ste10, p. 223] Find the derivative of the following functions

(i) (1 pt) y = (sin x)

ln x

(ii) (1 pt) y = (tan x)

1/x

(2 pts)

Exercise 3.7 [Ste10, p. 272] If

y =

x

√

a

2 − 1

−

2

√

a

2 − 1

arctan

sin x

a +

√

a

2 − 1 + cos x

Show that y

′ =

1

a + cos x

.

(2 pts)

Exercise 3.8 [Ste10, p. 282] If f has a local minimum value at c, show that the function g(x) = −f(x) has a local

maximum value at c.

(2 pts)

Exercise 3.9 [Ste10, p. 289] Suppose f is an odd function and is differentiable everywhere. Show that for every

positive number b, there exists a number c ∈ (−b, b) such that f

′

(c) = f(b)/b.

(2 pts)

Exercise 3.10 [Ste10, p. 300] Show that the inflection points of the curve y = x sin x lie on the curve y

2

(x

2+4) = 4x

2

.

(2 pts)

Exercise 3.11 [Ste10, p. 309] Evaluate

limx→∞

x − x

2

ln

1 + x

x

(2 pts)

Exercise 3.12 [Ste10, p. 309] Let

f(x) = (

|x|

x

, x ̸= 0

1, x = 0

(i) (1 pt) Show that f is continuous at 0.

(ii) (1 pt) Calculate f

′

(0).

(2 pts)

Exercise 3.13 [Ste10, p. 309] Show that the shortest distance from the point (x1, y1) to the straight line Ax + By + C = 0

is2

|Ax1 + By1 + C|

√

A2 + B2

(2 pts)

References

[Ste10] J. Stewart. Calculus: Early Transcendentals. 7th ed. Cengage Learning, 2010 (Cited on pages 1, 2).

2of course one approach is to use the Cauchy-Schwarz inequality.

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