Given that $y=x^5 + 1/(x^2)$, find

(i) $\frac{\text{d}y}{\text{d}x}$

(ii) $\frac{d^2y}{dx^2}$

$y=x^5 + 1/(x^2)$

or $y=x^5 + x^{-2}$

$\therefore \frac{\text{d}y}{\text{d}x}$ = $5x^4-2x^{-3}$

$\frac{\text{d}y}{\text{d}x}$ = $5x^4-2x^{-3}$
$\therefore \frac{d^2y}{dx^2}$ = $5\times4x^3-2\times(-3)x^{-4}$
or, $\frac{d^2y}{dx^2}$ = $20x^3+6x^{-4}$