MATH 1210 - Section 002

Fall 2010

Nicolas Fourrier

Due on September 30, 2010

Homework 5 Name:

1. Find the derivative of the function f by using the rules of differentation. Credits are given for the details. q  2 3 +1 f1 (x) = x−1 f2 (x) = xx2 −1 x+1 f3 (x) = (1 + 4x)5 (3 + x − x2 )8 f5 (x) = f7 (x) =

(x−1)4 (x2 +2x)5

q

x+

f10 (x) =

p

f4 (x) = f6 (x) =

x+

√

1 (t4 +1)3 q

x x2 +4

√ f8 (x) = x 2 − x2

x

1 1 (x−1) 2

3 (x+1) 2

f30 (x) = 4(1 + 4x)4 (3 + x − x2 )7 (17 + 9x − 21x2 ) f50 (x) =

3

2

2(x−1) (−3x +4x+5) x2 +2x)6

f70 (x) = 12 (x +

p

x+

√

x)

−1 2

f20 (x) =

−12x(x2 +1)2 (x2 −1)4

f40 (x) =

−12t3 (t4 +1)4 4−x2 3 √ 2 x(x2 +4) 2 2 2−2x √

f60 (x) = h i −1 √ −1 1 + 21 (x + x) 2 (1 + 12 x 2 )

f80 (x) =

2−x2

2. A particle moves along a straight line with displacement s(t), velocity v(t), and acceleration a(t). Show that a(t) = v(t) dv . Use the chain rule, and the fact that the acceleration is the derivative the velocity and the velocity ds is the derivative of the displacement. Explain what is dv . ds ds By the chain rule, a(t) = dv = dv = dv v(t). dt ds dt ds dv The derivative ds is the rate of change of the velocity with respect to the displacement.

3. The curve y = √ |x|

2−x2

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

point (1, 1) For x > 0, |x| = x, and y = f (x) = √ x 2 . 2−x √ 3 So f 0 (x) = 2(2 − x2 ) 2 . So at (1, 1), the slope of the tangent line is f 0 (1) = 2 and its equation is y = 2x − 1. 4. Suppose a company has estimated that the cosst (in dollars) of producing x items is C(x) = 10000+5x+0.01x2 . (a) What is the marginal cost function ? (b) What is the marginal cost at the production level of 500 items.

(a) The marginal cost function is C 0 (x) = 5 + 0.02x. (b) C 0 (500) = 5 + 0.02(500) = $15/item.

5. The cost, indollars, of producing x yards of a certain fabric is C(x) = 1200 + 12x − 0.01x2 + 0.0005x3 . (a) Find the marginal cost function. (b) Find C 0 (200) and explain its meaning. What does it predict ? (c) Compare C 0 (200) with the cost of manufacturing the 201st yard of fabric ?

(a) The marginal cost function is C 0 (x) = 12 − 0.2x + 0.0015x2 $/yard. (b) C 0 (x) = 12 − 0.2(200) + 0.0015(200)2 = 32$/yard. This is the rate at which costs are increasing with respect the production level when x = 200. C 0 (200) predicts the cost of producing the 201st yard. (c) The cost of manufacturing the 201st yard is C(201) − C(200) = $32.20, which is approximately C 0 (200)

6. Find the second derivative of the function f by using the rules of differentation. Credits are given for the details. f1 (x) = 339 + 25x − 0.09x2 + 0.0004x3 f2 (x) = x1 f3 (x) = 1 + 4x − x2 f4 (x) = sqrt(1 + x)(4 + 2x2 ) f100 (x) = −0.18 + 0.0024x f300 (x) = −2 f400 (x) = √8x+4(x+1) 2

(1+x)(4+2x2 )

(2x2 +4(x+1)x+4)2 3 4((x+1)(2x2 +4)) 2

f200 (x) = x23 2 √ +4(x+1)x+4 f40 (x) = 2x 2 2

(x+1)(2x +4)