<s>[INST] Simplify the difference quotient $\frac{f(x+h)-f(x)}{h}$. 
Simplify the difference quotient $\frac{f(x+h)-f(x)}{h}$ where
  a) $f(x)=2x+3,$
  b) $f(x)=\frac{1}{x+1},$
  c) $f(x)=x^2.$  

I believe that if anyone can help me out with the first one, the other two might come clearer to me. But I started out this problem by plugging in $f(x)$ and got: 
$$\dfrac{(2x+3)+f(h)-(2x+3)}{h}$$
I have no idea what to do after this. Because the $2x+3$'s can cancel out and leave me with just $\frac{f(h)}{h}$ but that doesn't make sense to me. Please help.
EDIT: The first one is solved and now for the second one, this is what I got:
$$\dfrac{\frac{1}{(x+h)+1}-\frac{1}{x+1}}{h}$$ Now, to subtract fractions the denominator has to be the same and they are the exact same except for the first fraction has an $h$ while the other one does not. How would I go about this?
For the third problem, this is all of my work: 
$$\begin{align*}\dfrac{(x+h)^2-x^2}{h}&=
\dfrac{x^2+2hx+h^2-x^2}{h}\\
&= \dfrac{2hx+h^2}{h}\\
&= \dfrac{h(2x+h)}{h}\\
&= 2x+h\end{align*}$$
Is this all correct?
 [/INST] $f(x+h)$ means replace $x$ with $x+h$ in your function definition. If $f(x) = 2x+3$, then $f(x+h) = 2(x+h)+3$, not $2x+3+f(h)$.
Therefore, $$\frac{f(x+h)-f(x)}{h} = \frac{2(x+h)+3-\left(2x+3\right)}{h} = \frac{2x+2h+3-2x-3}{h} = \cdots$$
Edit: To answer your second question, how do you handle just $\frac{1}{x+h+1}-\frac{1}{x+1}$? One simple way: multiply the first term above and below by the second term's denominator, and vice-versa: $$\frac{1}{x+h+1}\frac{x+1}{x+1}-\frac{1}{x+1}\frac{x+h+1}{x+h+1} = \frac{(x+1)-(x+h+1)}{(x+h+1)(x+1)}.$$
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