How do you find f'(x) using the limit definition given #f (x) = sqrt(1+3x)#?

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Answer 1 · 2016-07-04T18:18:59

#= 3 / (2sqrt(1+3x )#

#f (x) = sqrt(1+3x)#

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

#= lim_[h to 0] 1/h (sqrt(1+3(x+h)) - sqrt(1+3x))#

multiply by conjugate
#= lim_[h to 0] 1/h (sqrt(1+3(x+h)) - sqrt(1+3x)) times (sqrt(1+3(x+h)) + sqrt(1+3x))/(sqrt(1+3(x+h)) + sqrt(1+3x))#

#= lim_[h to 0] 1/h (1+3(x+h) - (1+3x)) / (sqrt(1+3(x+h)) + sqrt(1+3x))#

#= lim_[h to 0] 1/h (3h) / (sqrt(1+3(x+h)) + sqrt(1+3x))#

#= lim_[h to 0] (3) / (sqrt(1+3(x+h)) + sqrt(1+3x))#

#= (3) / (sqrt(1+3(x)) + sqrt(1+3x))#

#= 3 / (2sqrt(1+3x )#