Find the derivatives of the following

Question

i. y=6x^2+2x-2 ii. y=(3x-4x^3)/((x+1))

Answer 1

Remember the formula d/dx(x^n) = nx^n-1

1) Given y = 6x^2+2x-2

Apply derivative on each side.

y' = d/dx(6x^2+2x-2)

y' = 12x+2

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

Apply the formula d/dx(u/v) = (uv'-vu')/v^2

y' = [(3x-4x^3)d/dx(x+1)-(x+1)d/dx(3x-4x^3)]/(x+1)^2

y' = [(3x-4x^3)-(x+1)(3-12x^2)]/(x+1)^2

y' = [(3x-4x^3)-(3x-12x^3+3-12x^2)]/(x+1)^2

y' = [3x-4x^3-3x+12x^3-3+12x^2]/(x+1)^2

y' = (8x^3+12x^2)/(x+1)^2

Comments

Apply the formula d/dx(u/v) = (vu'-uv')/v^2

y' = [(x + 1)(3x - 4x^3) ' - (3x - 4x^3)(x + 1) '] / (x+1)^2

y' = [(x + 1)(3 - 12x^2) - (3x - 4x^3)(1)] / (x+1)^2

y' = [3x -12x^3 + 3 - 12x^2 - 3x + 4x^3] / (x+1)^2

y' = (- 8x^3 - 12x^2 + 3) / (x+1)^2