# (Calculus) alternative derivative formula

I'd like to calculate the derivative of:

$$f(x)=\frac{1}{(x+2)}$$

using the alternative formula for calculating derivatives:

$f'(a)=\lim_{z\to a}\frac{f(z)-f(a)}{z-a}$

In an attempt to solve the problem, I first wrote down:

$$\lim_{z\to a}\frac{\frac{1}{z+2}-\frac{1}{a+2}}{z-a}$$

In an attempt to find a common factor in the nominator and denominator, I fail. I'd like to find a way to split the denominator in a product which makes it possible to cancel out the z-a (=a-a because $\lim_{z\to a}$) factor in the denominator since you can't divide by zero.

Can anyone help me a step further?

## 2 Answers

$$\lim_{z\to a}\frac{\frac{1}{z+2}-\frac{1}{\color{red}{x}+2}}{z-a}$$

Why is there still an $x$? You seem to be mixing $x$ and $a$.

$$\frac{\frac{1}{z+2}-\frac{1}{\color{blue}{a}+2}}{z-a} = \frac{\frac{(a+2)-(z+2)}{(z+2)(a+2)}}{z-a} = \frac{\frac{-(z-a)}{(z+2)(a+2)}}{z-a} = \cdots$$ Now you can cancel the common factor $z-a$ and proceed with taking the limit.

• thanks for your step-by-step solution!!:) By the way, I can't find the x you mean in my question. – Coen Oct 4 '16 at 19:30
• It's no longer there, someone edited your question ;-). You're welcome! – StackTD Oct 4 '16 at 20:12
• Aah, that explains :D – Coen Oct 6 '16 at 13:48

Observe \begin{align} \frac{\frac{1}{z+2}-\frac{1}{a+2}}{z-a} = \frac{\frac{-(z-a)}{(z+2)(a+2)}}{z-a} = \frac{-1}{(z+2)(a+2)}. \end{align}

• Thanks for your solution!!:) – Coen Oct 4 '16 at 19:29