Here is how I did them originally:

Differentiate the following

2. Find the gradient of the curve $\displaystyle y = x^2 + \frac{1}{x}$ at the point (1,2)

$\displaystyle \frac{dy}{dx} = 2x+\bold{\color{red}(-1)}x^{\bold{\color{red}-2}}$

$\displaystyle =2(1) + \bold{\color{red}(-1)}(1)^{\bold{\color{red}-2}}$

$\displaystyle =2\bold{\color{red}-1}$

$\displaystyle = \bold{\color{red}1}$

3. Differentiate

(i) $\displaystyle y=(x+2)(x^3+5x)$

$\displaystyle =(\bold{\color{red}x}+2)(\bold{\color{red}3x^2}+5) +(x^3+5x)(1)$

**<<<<<< here you calculate $\displaystyle 1 \cdot 3 = 4$. Quite unusual **
$\displaystyle =\bold{\color{red}4x^3}+5x-6x^2+10+x^3+5x$

$\displaystyle =5x^3 + 10x - 6x^2 +10$

(ii) $\displaystyle y=\frac{3x+2}{2x+1}$

$\displaystyle =\frac{dy}{dx} = 3x+2$

$\displaystyle =\frac{dy}{dx} = 3$

$\displaystyle =\frac{dy}{dx} = 2x+1$

$\displaystyle =\frac{dy}{dx} = 2$

$\displaystyle =\frac{(2x+1)3 - (3x+2)2}{(2x+1)^2}$

$\displaystyle =\frac{(6x+3-6x\bold{\color{red}-}4)}{(2x+1)^2}$

$\displaystyle =\frac{\bold{\color{red}-1}}{(2x+1)^2}$

I did this after your help, $\displaystyle \frac{3}{x^2}$

I got:

$\displaystyle = 3 \cdot x^{-2}$

**<<<<<<<< The number 3 is a constant factor.**
$\displaystyle = \bold{\color{red}3 \cdot (-2)}(x)^{-3}$

...

Thanks for the help so far.