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

Question

Isabella Ackerman · Accepted Answer

#-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)#

Ezekiel Alexander · Answer

undefined To find $ f'(x) $ using the definition of a derivative for $ f(x) = \frac{{2 - x}}{{2 + x}} $, follow these steps:

1. Use the definition of the derivative:
$$ f'(x) = \lim_{{h \to 0}} \frac{{f(x + h) - f(x)}}{h} $$

2. Substitute $ f(x) $ into the formula:
$$ f'(x) = \lim_{{h \to 0}} \frac{{\frac{{2 - (x + h)}}{{2 + (x + h)}} - \frac{{2 - x}}{{2 + x}}}}{h} $$

3. Simplify the expression inside the limit:
$$ f'(x) = \lim_{{h \to 0}} \frac{{\frac{{2 - x - h}}{{2 + x + h}} - \frac{{2 - x}}{{2 + x}}}}{h} $$

4. Combine the fractions:
$$ f'(x) = \lim_{{h \to 0}} \frac{{(2 - x - h)(2 + x) - (2 - x)(2 + x + h)}}{{h(2 + x)(2 + x + h)}} $$

5. Expand and simplify the numerator:
$$ f'(x) = \lim_{{h \to 0}} \frac{{4 - 2x + 2x - x^2 - 2h - xh - 4 + 2x + 2x^2 + 2x + xh - 2x - x^2 - 2h - xh}}{{h(2 + x)(2 + x + h)}} $$

6. Cancel out terms:
$$ f'(x) = \lim_{{h \to 0}} \frac{{-2h - 2h}}{{h(2 + x)(2 + x + h)}} $$

7. Simplify further:
$$ f'(x) = \lim_{{h \to 0}} \frac{{-4h}}{{h(2 + x)(2 + x + h)}} $$

8. Cancel out $ h $:
$$ f'(x) = \lim_{{h \to 0}} \frac{{-4}}{{(2 + x)(2 + x + h)}} $$

9. Evaluate the limit as $ h $ approaches 0:
$$ f'(x) = \frac{{-4}}{{(2 + x)(2 + x)}} $$

10. Simplify the expression:
$$ f'(x) = \frac{{-4}}{{(2 + x)^2}} $$

So, $ f'(x) = \frac{{-4}}{{(2 + x)^2}} $.