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

Question

1220 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-10-20T11:42:11

#-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)#