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

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Answer 1 · 2016-07-04T18:18:59

$= \frac{3}{2 \sqrt{1 + 3 x}}$ $= 3 / (2sqrt(1+3x )$

$f (x) = \sqrt{1 + 3 x}$ $f (x) = sqrt(1+3x)$

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

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

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

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

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

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

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