How do you find f'(x) using the limit definition given #sqrt(2x-1)#?

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Answer 1 · 2016-07-14T21:36:46

#f'(x) = lim_{h to 0} (f(x+h) - f(x))/(h)#

#=lim_{h to 0} (sqrt(2(x+h)-1) - sqrt(2x-1))/(h)#

next, we times by the conjugate

# =lim_{h to 0} (sqrt(2(x+h)-1) - sqrt(2x-1))/(h) * (sqrt(2(x+h)-1) + sqrt(2x-1))/(sqrt(2(x+h)-1) + sqrt(2x-1))#

#= lim_{h to 0} (1/h) (2(x+h)-1 - (2x-1))/(sqrt(2(x+h)-1) + sqrt(2x-1)) #

#= lim_{h to 0} (1/h) (2h)/(sqrt(2(x+h)-1) + sqrt(2x-1)) #

#= lim_{h to 0} (2)/(sqrt(2(x+h)-1) + sqrt(2x-1)) #

#= (2)/(sqrt(2x-1) + sqrt(2x-1)) #

#= (2)/(2(sqrt(2x-1) ) #

#= 1/(sqrt(2x-1) #