Last years midterm

Q1B:

find derivative with respect to x

y = (4x+3)^1/2 / 2x

u = (4x+3)

u' = 4

T'B-TB'/B^2

1/2(4x+3)-^1/2 * 4

2(4x+3)^-1/2

(2(4x+3)-^1/2 * 2x - (4x+3)^1/2 * 2) / (4x^2)

=

(4x(4x+3)^-1/2 - 2(4x+3)^1/2) / (4x^2)

(however, this derivative is in respect to y. Can someone still tell me if the derivative is correct at least, thanks)

Q2:

Find equation of line normal to

y = 2x^3 - 3x - 1 at x = 1

y - y1 = m(x - x1)

y' = m

y' = 6x^2 - 3

y' = 6(1)^2 - 3 = 36 - 3 = 33

y = 2(1)^3 - 3(1) - 1 = 8 - 3 - 1 = 4

x = 1, y = 4

y - 4 = 33(x - 1)

y - 4 = 33x - 33

y - 37 = 33x

Q4:

A rectangular area divided into two areas, wall inside parallel to 2 opposite sides

Find max dimensions for display if total fencing available is 240 m.

(sketch: 4 sides with one partitioning wall)

2x + 3y

240 = 2x + 3y

240 - 2x = 3y

80 - 2/3x = y

A = x * y

A = x * (80 - 2/3x)

A = 80x - 2/3x^2

A' = 80 - 4/3x

x = 60 m

y = 40 m

Q6:

Evaluate integrals:

A:

(x3 - x^1/2) / x

(x^3)/x - (x^1/2)/x

x^2 - x-^1/2

(1/3x^3) - (2x^1/2)

B:

8x(2x^2-5)^4

(2x^2-5)(2x^2-5)(2x^2-5)(2x^2-5)

4x^4 - 10x^2 - 10x^2 + 25

4x^4 - 20x^2 + 25

integrate

16x^8 - 80x^6 + 100x^4 - 80x^2 + 400x^4 + 500 + 100x^4 - 500x^2 + 625

16x^8 - 80x^6 + 500x^4 - 580x^2 + 1125

simplify

16/9x^9 - 80/7x^7 + 100x^5 + 580/3x^3 + 1125x

Q7:

Find area under curve

y = 1/(x + 1)^1/2

Between x = 3 and x = 8

(x + 1)^-1/2

2(x+1)^1/2

x = 3: 2(3+1)^1/2: y = 4

x = 8: 2(8+1)^1/2: y = 6

4 - 6 = -2

Thanks for looking at it.