How do you use the definition of a derivative to find the derivative of #f(x) = sqrtx + 2# to calculate f'(2)?

Question

1222 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-05T15:57:11

#[d/dx (sqrt(x) + 2)]_(x=2) = 1/(2sqrt2)#

By definition of derivative:

#f'(2) = lim_(h->0) (f(2+h)-f(2))/h#

For #f(x) = sqrt(x) + 2# this is:

#f'(2) = lim_(h->0) (sqrt(2+h) + 2 -sqrt(2) -2)/h#

simplify:

#f'(2)= lim_(h->0) (sqrt(2+h) -sqrt(2) )/h#

rationalize the numerator using the identity #(a+b)(a-b) = (a^2-b^2)#:

#f'(2)= lim_(h->0) ((sqrt(2+h) -sqrt(2) )/h )( (sqrt(2+h) + sqrt(2) )/(sqrt(2+h) +sqrt(2) ))#

#f'(2) = lim_(h->0) (2+h -2) /(h (sqrt(2+h) +sqrt(2) ))#

#f'(2) = lim_(h->0) h /(h (sqrt(2+h) +sqrt(2) ))#

#f'(2) = lim_(h->0) 1 / (sqrt(2+h) +sqrt(2) )#

#f'(2) = 1/(2sqrt(2))#