How do you find f'(x) using the limit definition given #4/(sqrt(x))#?

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Answer 1 · 2016-07-01T20:22:21

#= -2/(x^(3/2))#

by definition

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

here

#f'(x) = lim_{h to 0} (4/sqrt(x +h) - 4/sqrtx)/(h)#

multiply by the conjugate

#f'(x) = lim_{h to 0} (4/sqrt(x +h) - 4/sqrtx)/(h) * (4/sqrt(x +h) + 4/sqrtx)/ (4/sqrt(x +h) + 4/sqrtx)#

#f'(x) = lim_{h to 0} (16/(x +h) - 16/x)/(h) * (1)/ (4/sqrt(x +h) + 4/sqrtx)#

start to combine terms

#f'(x) = 16 lim_{h to 0} ((x - x - h)/(x(x +h)))/(h) * (1)/ (4 (sqrt(x+h) + sqrtx)/(sqrt(x +h) sqrtx))#

#f'(x) = 4 lim_{h to 0} (-h)/(h*x(x+h)) * ((sqrt(x +h) sqrtx))/ ( (sqrt(x+h) + sqrtx))#

#f'(x) = -4 lim_{h to 0} (1)/(sqrt(x +h) sqrtx) * (1)/ ( (sqrt(x+h) + sqrtx))#

setting h = 0

#f'(x) = -4(1)/(sqrt(x ) sqrtx) * (1)/ ( (sqrt(x) + sqrtx))#

#= -4(1)/(x) * (1)/ ( 2sqrt(x) )#

#= -2/(x^(3/2))#