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

To find the first and second derivatives for each function, I will apply standard differentiation rules: the power rule, constant multiple rule, sum/difference rule, quotient rule, and chain rule where necessary.

Step-by-Step Derivations:

a) $y = x^2 + 2x - 12$
* First Derivative ($y'$):
* $\frac{d}{dx}(x^2) = 2x$
* $\frac{d}{dx}(2x) = 2$
* $\frac{d}{dx}(-12) = 0$
* $y' = 2x + 2$
* Second Derivative ($y''$):
* $\frac{d}{dx}(2x) = 2$
* $\frac{d}{dx}(2) = 0$
* $y'' = 2$

b) $f(x) = 4x - 5$
* First Derivative ($f'(x)$):
* $\frac{d}{dx}(4x) = 4$
* $\frac{d}{dx}(-5) = 0$
* $f'(x) = 4$
* Second Derivative ($f''(x)$):
* The derivative of a constant (4) is 0.
* $f''(x) = 0$

c) $y = x^3 - 4x^2 + 8x - 7$
* First Derivative ($y'$):
* $\frac{d}{dx}(x^3) = 3x^2$
* $\frac{d}{dx}(-4x^2) = -8x$
* $\frac{d}{dx}(8x) = 8$
* $\frac{d}{dx}(-7) = 0$
* $y' = 3x^2 - 8x + 8$
* Second Derivative ($y''$):
* $\frac{d}{dx}(3x^2) = 6x$
* $\frac{d}{dx}(-8x) = -8$
* $\frac{d}{dx}(8) = 0$
* $y'' = 6x - 8$

d) $y = \sqrt{x} + \frac{1}{2x}$
* Rewrite using exponents: $y = x^{1/2} + \frac{1}{2}x^{-1}$
* First Derivative ($y'$):
* $\frac{d}{dx}(x^{1/2}) = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$
* $\frac{d}{dx}(\frac{1}{2}x^{-1}) = \frac{1}{2}(-1)x^{-2} = -\frac{1}{2x^2}$
* $y' = \frac{1}{2\sqrt{x}} - \frac{1}{2x^2}$
* Second Derivative ($y''$):
* $\frac{d}{dx}(\frac{1}{2}x^{-1/2}) = \frac{1}{2}(-\frac{1}{2})x^{-3/2} = -\frac{1}{4}x^{-3/2} = -\frac{1}{4x\sqrt{x}}$
* $\frac{d}{dx}(-\frac{1}{2}x^{-2}) = -\frac{1}{2}(-2)x^{-3} = x^{-3} = \frac{1}{x^3}$
* $y'' = -\frac{1}{4x\sqrt{x}} + \frac{1}{x^3}$

e) $f(x) = \frac{6}{x} - 11x^2$
* Rewrite: $f(x) = 6x^{-1} - 11x^2$
* First Derivative ($f'(x)$):
* $\frac{d}{dx}(6x^{-1}) = -6x^{-2} = -\frac{6}{x^2}$
* $\frac{d}{dx}(-11x^2) = -22x$
* $f'(x) = -\frac{6}{x^2} - 22x$
* Second Derivative ($f''(x)$):
* $\frac{d}{dx}(-6x^{-2}) = 12x^{-3} = \frac{12}{x^3}$
* $\frac{d}{dx}(-22x) = -22$
* $f''(x) = \frac{12}{x^3} - 22$

f) $g(x) = \frac{x^2+9}{x+1}$
* Use Quotient Rule: $(\frac{u}{v})' = \frac{u'v - uv'}{v^2}$
* $u = x^2+9 \Rightarrow u' = 2x$
* $v = x+1 \Rightarrow v' = 1$
* First Derivative ($g'(x)$):
* $g'(x) = \frac{(2x)(x+1) - (x^2+9)(1)}{(x+1)^2}$
* $g'(x) = \frac{2x^2 + 2x - x^2 - 9}{(x+1)^2}$
* $g'(x) = \frac{x^2 + 2x - 9}{(x+1)^2}$
* Second Derivative ($g''(x)$):
* Apply Quotient Rule again to $g'(x)$.
* $U = x^2 + 2x - 9 \Rightarrow U' = 2x + 2$
* $V = (x+1)^2 \Rightarrow V' = 2(x+1)$
* $g''(x) = \frac{(2x+2)(x+1)^2 - (x^2+2x-9)(2(x+1))}{((x+1)^2)^2}$
* Factor out $2(x+1)$ from numerator:
* $g''(x) = \frac{2(x+1)[(x+1)^2 - (x^2+2x-9)]}{(x+1)^4}$
* Simplify inside brackets: $(x^2+2x+1) - (x^2+2x-9) = 10$
* $g''(x) = \frac{2(x+1)(10)}{(x+1)^4}$
* $g''(x) = \frac{20}{(x+1)^3}$

g) $y = x^5 + 7x^2 - 12$
* First Derivative ($y'$):
* $\frac{d}{dx}(x^5) = 5x^4$
* $\frac{d}{dx}(7x^2) = 14x$
* $y' = 5x^4 + 14x$
* Second Derivative ($y''$):
* $\frac{d}{dx}(5x^4) = 20x^3$
* $\frac{d}{dx}(14x) = 14$
* $y'' = 20x^3 + 14$

h) $y = -x^3 + 4x^2 - 18x - 2$
* First Derivative ($y'$):
* $\frac{d}{dx}(-x^3) = -3x^2$
* $\frac{d}{dx}(4x^2) = 8x$
* $\frac{d}{dx}(-18x) = -18$
* $y' = -3x^2 + 8x - 18$
* Second Derivative ($y''$):
* $\frac{d}{dx}(-3x^2) = -6x$
* $\frac{d}{dx}(8x) = 8$
* $y'' = -6x + 8$

i) $h(x) = \frac{3}{x^3} - \sqrt{x^3}$
* Rewrite: $h(x) = 3x^{-3} - x^{3/2}$
* First Derivative ($h'(x)$):
* $\frac{d}{dx}(3x^{-3}) = -9x^{-4} = -\frac{9}{x^4}$
* $\frac{d}{dx}(-x^{3/2}) = -\frac{3}{2}x^{1/2} = -\frac{3\sqrt{x}}{2}$
* $h'(x) = -\frac{9}{x^4} - \frac{3\sqrt{x}}{2}$
* Second Derivative ($h''(x)$):
* $\frac{d}{dx}(-9x^{-4}) = 36x^{-5} = \frac{36}{x^5}$
* $\frac{d}{dx}(-\frac{3}{2}x^{1/2}) = -\frac{3}{2}(\frac{1}{2})x^{-1/2} = -\frac{3}{4\sqrt{x}}$
* $h''(x) = \frac{36}{x^5} - \frac{3}{4\sqrt{x}}$

j) $y = \frac{1}{2}x^3 + 2x^2 - 3x + 2$
* First Derivative ($y'$):
* $\frac{d}{dx}(\frac{1}{2}x^3) = \frac{3}{2}x^2$
* $\frac{d}{dx}(2x^2) = 4x$
* $\frac{d}{dx}(-3x) = -3$
* $y' = \frac{3}{2}x^2 + 4x - 3$
* Second Derivative ($y''$):
* $\frac{d}{dx}(\frac{3}{2}x^2) = 3x$
* $\frac{d}{dx}(4x) = 4$
* $y'' = 3x + 4$

Final Answer:
a) $y' = 2x + 2$, $\quad y'' = 2$
b) $f'(x) = 4$, $\quad f''(x) = 0$
c) $y' = 3x^2 - 8x + 8$, $\quad y'' = 6x - 8$
d) $y' = \frac{1}{2\sqrt{x}} - \frac{1}{2x^2}$, $\quad y'' = -\frac{1}{4x\sqrt{x}} + \frac{1}{x^3}$
e) $f'(x) = -\frac{6}{x^2} - 22x$, $\quad f''(x) = \frac{12}{x^3} - 22$
f) $g'(x) = \frac{x^2 + 2x - 9}{(x+1)^2}$, $\quad g''(x) = \frac{20}{(x+1)^3}$
g) $y' = 5x^4 + 14x$, $\quad y'' = 20x^3 + 14$
h) $y' = -3x^2 + 8x - 18$, $\quad y'' = -6x + 8$
i) $h'(x) = -\frac{9}{x^4} - \frac{3\sqrt{x}}{2}$, $\quad h''(x) = \frac{36}{x^5} - \frac{3}{4\sqrt{x}}$
j) $y' = \frac{3}{2}x^2 + 4x - 3$, $\quad y'' = 3x + 4$