How do you use the definition of a derivative to find f' given #f(x)=sqrtx# at x=1?

1 Answer
Steve M
Nov 3, 2016

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

Explanation:

By definition # f'(x) =lim_(hrarr0)( (f(x+h)-f(x))/h ) #

So, with # f(x)=sqrt(x) # we have:
# f'(x) = lim_(hrarr0)( sqrt(x+h) - sqrt(x) ) / h #
# :. f'(x) = lim_(hrarr0)( (sqrt(x+h) - sqrt(x) )) / h *( (sqrt(x+h) + sqrt(x) ))/(( sqrt(x+h) + sqrt(x) ))#
# :. f'(x) = lim_(hrarr0) (sqrt(x+h)sqrt(x+h)-sqrt(x)sqrt(x))/(h( sqrt(x+h) + sqrt(x) ))#
# :. f'(x) = lim_(hrarr0) (x+h-x)/(h( sqrt(x+h) + sqrt(x) ))#
# :. f'(x) = lim_(hrarr0) h/(h( sqrt(x+h) + sqrt(x) ))#
# :. f'(x) = lim_(hrarr0) 1/( sqrt(x+h) + sqrt(x) )#
# :. f'(x) = 1/( sqrt(x) + sqrt(x) )#
# :. f'(x) = 1/( 2sqrt(x) )#

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