How do I use the limit definition of derivative to find #f'(x)# for #f(x)=sqrt(2+6x)# ?

Question

6526 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2014-10-03T04:01:33

#f'(x)=lim_(h->0) (f(x+h)-f(x))/h#

#f(x)=sqrt(2+6x)#

#f(x+h)=sqrt(2+6(x+h))=sqrt(2+6x+6h#

Make the substitutions for #f(x)# and #f(x+h)#

#f'(x)=lim_(h->0) (sqrt(2+6x+6h)-sqrt(2+6x))/h#

Rationalize the numerator

#=lim_(h->0) (sqrt(2+6x+6h)-sqrt(2+6x))/h*(sqrt(2+6x+6h)+sqrt(2+6x))/(sqrt(2+6x+6h)+sqrt(2+6x))#

Remember the difference of perfect squares for the numerator

#=lim_(h->0) ((2+6x+6h)-(2+6x))/(h*sqrt(2+6x+6h)+sqrt(2+6x))#

Distribute the negative

#=lim_(h->0) (2+6x+6h-2-6x)/(h*sqrt(2+6x+6h)+sqrt(2+6x))#

Simplify numerator

#=lim_(h->0) (6h)/(h*sqrt(2+6x+6h)+sqrt(2+6x))#

Cancel the factors of #h#

#=lim_(h->0) (6)/(sqrt(2+6x+6h)+sqrt(2+6x))#

Substitute in the value of 0 for #h# and then simplify

#=(6)/(sqrt(2+6x+6(0))+sqrt(2+6x))#

#=(6)/(sqrt(2+6x)+sqrt(2+6x))#

#=(6)/(2sqrt(2+6x))#

#=3/sqrt(2+6x)#