Grade 12 Mathematics worksheet focusing on calculus, specifically finding first and second derivatives of various algebraic functions.

a) y = x³ + 3x - 12
First derivative: y' = 3x² + 3
Second derivative: y'' = 6x

b) f(x) = 4x - 5
First derivative: f'(x) = 4
Second derivative: f''(x) = 0

c) y = x⁷ - 4x⁴ + 8x - 7
First derivative: y' = 7x⁶ - 16x³ + 8
Second derivative: y'' = 42x⁵ - 48x²

d) y = √x + 1/x²
Rewrite: y = x^(1/2) + x^(-2)
First derivative: y' = (1/2)x^(-1/2) - 2x^(-3)
Second derivative: y'' = (-1/4)x^(-3/2) + 6x^(-4)

e) f(x) = 5/x - 11x²
Rewrite: f(x) = 5x^(-1) - 11x²
First derivative: f'(x) = -5x^(-2) - 22x
Second derivative: f''(x) = 10x^(-3) - 22

f) g(x) = (x² - 5)/(x + 1)
Use quotient rule: g'(x) = [(2x)(x+1) - (x²-5)(1)] / (x+1)² = (2x² + 2x - x² + 5) / (x+1)² = (x² + 2x + 5) / (x+1)²
Second derivative: Use quotient rule again on g'(x):
Let u = x² + 2x + 5, v = (x+1)²
u' = 2x + 2, v' = 2(x+1)
g''(x) = [u'v - uv'] / v² = [(2x+2)(x+1)² - (x²+2x+5)(2)(x+1)] / (x+1)⁴
Factor (x+1) in numerator: = (x+1)[(2x+2)(x+1) - 2(x²+2x+5)] / (x+1)⁴
Simplify inside: (2x+2)(x+1) = 2x² + 4x + 2; 2(x²+2x+5) = 2x² + 4x + 10
So: (2x² + 4x + 2 - 2x² - 4x - 10) = -8
Thus: g''(x) = (x+1)(-8) / (x+1)⁴ = -8 / (x+1)³

g) y = x⁴ + 7x² - 12
First derivative: y' = 4x³ + 14x
Second derivative: y'' = 12x² + 14

h) h(x) = 4/x³ - √x⁵
Rewrite: h(x) = 4x^(-3) - x^(5/2)
First derivative: h'(x) = -12x^(-4) - (5/2)x^(3/2)
Second derivative: h''(x) = 48x^(-5) - (15/4)x^(1/2)

i) y = -x⁵ + 4x² - 18x - 2
First derivative: y' = -5x⁴ + 8x - 18
Second derivative: y'' = -20x³ + 8

j) y = (1/5)x⁵ + (1/4)x⁴ - 3x + 2
First derivative: y' = x⁴ + x³ - 3
Second derivative: y'' = 4x³ + 3x²