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

1 Answer
Jul 1, 2016

= -2/(x^(3/2))=2x32

Explanation:

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))