How do you find f'(x) using the definition of a derivative for $f (x) = \frac{2 - x}{2 + x}$ $f(x)=(2-x)/(2+x)$ ?

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Answer 1 · 2015-10-20T11:42:11

$- \frac{4}{x^{2} + 4 x + 4}$ $-4/(x^2+4 x+4)$

The definition is the following:

$f'(x)=lim_{h->0} {f(x+h)-f(x)}/h$ .

So, $f(x)=(2-x)/(2+x)$ , and

$f(x+h)=(2-(x+h))/(2+(x+h))=(2-x-h)/(2+x+h)$ .

Let's compute $f(x+h)-f(x)$ :

$(2-x-h)/(2+x+h) - (2-x)/(2+x) = ((2-x-h)(2+x)-(2-x)(2+x+h))/((2+x)(2+x+h))$

Expanding both numerator and denominator, we get

$(-4h)/(h x+2 h+x^2+4 x+4)$

Dividing by $h$ , we have

$-4/(h x+2 h+x^2+4 x+4)$

Considering the limit as $h->0$ means to cancel the terms involving $h$ , so the result is

$-4/(x^2+4 x+4)$

How do you find f'(x) using the definition of a derivative for f(x)=(2-x)/(2+x) f(x)=2−x2+xf(x)=(2-x)/(2+x) ?