I don't understand how -12 is the numerator in this difference quotient. Find the difference quotient of $$\frac{\left(\frac{1}{12(x+h)+3}\right)-\left(\frac{1}{12x+3}\right)}{h}$$
When I solve it I end up canceling h's at the end and only having -12 as the answer. However, the correct answer is -12 over the denominator of the fractions in the numerator.
I solved it like this:
$$\frac{\left(\frac{1}{12x+12h+3}\right)-\left(\frac{1}{12x+3}\right)}{h}$$
Then multiplied each fraction by $(12x+12h+3)(12x+3)$ to get:
$$\frac{12x+3-12x-12h-3}{h}$$
Then simplify the numerator to this:
$$\frac{-12h}{h}$$
Then cancel h which gave me -12.

$$
\frac{\frac{1}{12(x+h)+3}-\frac{1}{12x+3}}{h}=\frac{1}{h}\cdot \left(\frac{1}{12(x+h)+3}-\frac{1}{12x+3}\right)=\\
\frac{1}{h}\cdot \frac{12x+3-12(x+h)-3}{(12(x+h)+3)(12x+3)}=\\
\frac{1}{h}\cdot \frac{-12h}{(12(x+h)+3)(12x+3)}=\frac{-12}{(12(x+h)+3)(12x+3)}
$$
