How do you find the derivative of #f(x) = x/(x+4) # using the limit definition?

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Answer 1 · 2016-06-30T12:46:22

#f'(x) = 4/(x+4)^2#

Limit definition of a derivative is:

#f'(x) = lim_(h->0) (f(x+h) - f(x))/h#

#f(x+h) = (x+h)/(x+h+4)#

#f(x) = x/(x+4)#

#f'(x) = lim_(h->0) ((x+h)/(x+h+4) - x/(x+4))/h#

Need to get a common denominator so:

#f'(x) = lim_(h->0) ((x+h)(x+4) - x(x+h+4))/((x+4)(x+h+4)) * 1/h#

#= lim_(h->0) (x^2 + (4+h)x + 4h - x^2 -(4+h)x)/((x+4)(x+h+4)) * 1/h#

A lot of the numerator cancels:

#= lim_(h->0) (cancel(x^2) + cancel((4+h)x) + 4h - cancel(x^2) -cancel((4+h)x))/((x+4)(x+h+4)) * 1/h#

We are left with:

# = lim_(h->0) 4/((x+4)(x+h+4)) = 4/(x+4)^2 #