First derivative of $f(x)= \frac{2}{x+1} +3$

Question

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}

Answer 1

In case you want to go by this definition.

In some cases it is more straightforward than the one you have used. If you put $x-a=h$ in the definition below you get the one you have.

There is an equivalent definition which says that function is differentiable at a point $a$ if $$\lim_{x\rightarrow a} \frac{f(x)-f(a)}{x-a}$$ exists and we denote this limit as $f'(a)$

$$\lim_{x\rightarrow a}\frac{\frac{2}{x+1}+3-\frac{2}{a+1}-3}{x-a}$$

$$\lim_{x\rightarrow a}\frac{\frac{2}{x+1}-\frac{2}{a+1}}{x-a}$$

$$\lim_{x\rightarrow a}\frac{\frac{-2(x-a)}{(a+1)(x+1)}}{x-a}$$

$$\lim_{x\rightarrow a}\frac{-2}{(a+1)(x+1)}$$

$$f'(a)=\frac{-2}{(a+1)(a+1)}=\frac{-2}{(a+1)^2}$$

Answer 2

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.