challenging calculus q?

Question

13. If y = 3x² +2x / 2x² -3 , calculate dy/dx.

a. – (2x² + 9x + 3) / ( 2x² -3)²
b. -2(2x² + 9x +3) / ( 2x²-3)²
c .-3(2x² -9x -3) / ( 2x² -3)²
d. -4( 2x² - 9x – 3) / (2x² - 3)²

Answer 1

This is a straight application of the quotient rule.
If y = u/v where u and v are functions of x then
dy/dx = (v*(du/dx) - u*(dv/dx))/v^2
In this case u = 3x^2 + 2x and v = 2x^2 - 3
You should be able to finish it yourself.

Answer 2

y = 3x² +2x / 2x² -3, therefore let u = 3x² +2x and v = 2x² -3. Then dy/dx = [v(du/dx) - u(dv/dx)]\v^2.

dy/dx = [(2x² -3)(6x+2) - (3x² +2x)(4x)]/(2x² - 3)².
= (12x^3+4x² - 18x - 6 - 12x^3 -8x²)/(2x² - 3)².
= (-4x² - 18x -6)/ (2x² - 3)².
dy/dx = -2(2x² + 9x +3)/(2x² - 3)². (b.)

Answer 3

d/dx (f(x)/g(x) = [f'(x)g(x) - f(x)g'(x)]/g(x)^2

dy/dx = [(6x + 2)(2x^2 - 3) - (3x^2 + 2x)*4x]/(2x^2 - 3)^2 =
= (12x^3 + 4x^2 - 18x - 6 - 12x^3 - 8x^2)/(2x^2 - 3)^2 =
= (-4x^2 - 18x - 6)/(2x^2 - 3)^2 = -2(x^2 + 9x + 3)/(2x^2 - 3)^2

The answer is b.

Answer 4

y = (3x² +2x) / (2x² -3) ; quotient rule

dy/dx = [(2x² -3)(6x + 2) - (4x)((3x² +2x)] / (2x² -3)²

dy/dx = [(12x^3 + 4x² - 18x - 6) - (12x^3 + 8x²)] / (2x² -3)²

dy/dx = (-4x² - 18x - 6) / (2x² -3)²

dy/dx = -2(2x² +x + 3) / (2x² -3)²

Answer: B

Answer 5

2) lim x->3 [(x-1)^2*(x-3)]/[(x+1)^2*(x-3)] = lim x->3 [(x-1)^2]/[(x+1)^2] = [(3-1)^2]/[(3+1)^2] =4/16=1/4 3) lim x->2 [√(3x-2)-√(x+2)]/(x-2) = lim x->2 [√(3x-2)-√(x+2)]/(x-2) * [√(3x-2)+√(x+2)]/[√(3x-2)+√(x+2)] = lim x->2 [(3x-2)-(x+2)]/(x-2)[√(3x-2)+√(x+2)] = lim x->2 [2(x-2)]/(x-2)[√(3x-2)+√(x+2)] = lim x->2 2/[√(3x-2)+√(x+2)] = 2/[√(6-2)+√(2+2)] = 2/4=1/2 4) lim q->0 [1-cos2q]/[4q^2] = lim q->0 [1-(1-sin^2 q)]/[4q^2] = lim q->0 (sin^2 q)/4q^2 = 1/4 6) lim x->∞ (x-(1/x))^2 = lim x->∞ x^2 - 2 - (1/x)^2 = 1 sorry that's all i can do

Answer 6

the right answer is no.(b)