<s>[INST] First derivative of $f(x)= \frac{2}{x+1} +3$ I'm struggling to find the first derivative of $f(x)= \frac{2}{x+1} +3$ using the limit definition of derivative. I keep coming up with $f'(x) = \frac{-5}
{(x+1)^2}$ but I should be getting $f'(x) = \frac{-2}{(x+1)^2}$
\begin{align}
\lim_{x\to 0} &= \frac{(\frac{2}{x+h+1}+3)-(\frac{2}{x+1}+3)}{h}\\
\lim_{x\to 0} &= \frac{\frac{2(x+1)+3(x+1)-(2(x+h+1)+3(x+h+1))}{(x+h+1)(x+1)}}{h}\\
\lim_{x\to 0} &= \frac{\frac{2x+2+3x+3-2x-2h-2-3x-3h-3}{(x+h+1)(x+1)}}{h}\\
\lim_{x\to 0} &= \frac{\frac{-5h}{(x+h+1)(x+1)}}{h}\\
\lim_{x\to 0} &= (\frac{-5h}{(x+h+1)(x+1)})*\frac{1}{h}\\
\lim_{x\to 0 }&= \frac{-5}{(x+1)^2}\\
\end{align}
 [/INST] The $3$ in the numerator should cancel out. 
Even if you want to keep it for a while, it shouldn't be $3(x+1)$ and $3(x+h+1)$, it should be $3(x+1)(x+h+1)$ for both of them.
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