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

Question

4625 views around the world

You can reuse this answer
Creative Commons License

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))#