How do you find the derivative of #sqrt (1+3x) using the limit definition?

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Answer 1 · 2016-07-30T23:14:09

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

$f'(x) = lim_(hrarr0) (f(x+h) - f(x))/h$

$=lim_(hrarr0) (sqrt(1+3(x+h)) - sqrt(1+3x))/h$

Multiply numerator and denominator by $sqrt(1+3(x+h)) + sqrt(1+3x)$

$lim_(hrarr0) (sqrt(1+3(x+h)) - sqrt(1+3x))/h(sqrt(1+3(x+h)) + sqrt(1+3x))/(sqrt(1+3(x+h)) + sqrt(1+3x))$

$lim_(hrarr0) (1+3(x+h) - (1+3x))/(h(sqrt(1+3(x+h)) + sqrt(1+3x)))$

$lim_(hrarr0) (3h)/(h(sqrt(1+3(x+h)) + sqrt(1+3x)))$

$lim_(hrarr0) 3/(sqrt(1+3(x+h)) + sqrt(1+3x)) = 3/(2sqrt(1+3x))$