How do you use the limit definition of the derivative to find the derivative of #f(x)=sqrt(4x-5)#?

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Answer 1 · 2017-02-28T01:15:40

#d/dx( sqrt(4x-5)) = 2/ sqrt(4x-5)#

By definition of derivative we have:

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

so:

#d/dx( sqrt(4x-5)) = lim_(h->0) (sqrt(4(x+h)-5)-sqrt(4x-5))/h#

rationalize the numerator using the identity:

#a^2-b^2 =(a-b)(a+b)#

#d/dx( sqrt(4x-5)) = lim_(h->0) ((sqrt(4(x+h)-5)-sqrt(4x-5))/h)( (sqrt(4(x+h)-5)+sqrt(4x-5))/(sqrt(4(x+h)-5)+sqrt(4x-5)))#

#d/dx( sqrt(4x-5)) = lim_(h->0) (4(x+h)-5-(4x-5))/(h (sqrt(4(x+h)-5)+sqrt(4x-5))#

#d/dx( sqrt(4x-5)) = lim_(h->0) (cancel(4x)+4h-cancel5-cancel(4x)+cancel5)/(h (sqrt(4(x+h)-5)+sqrt(4x-5))#

#d/dx( sqrt(4x-5)) = lim_(h->0) (4cancelh)/(cancelh (sqrt(4(x+h)-5)+sqrt(4x-5))#

#d/dx( sqrt(4x-5)) = lim_(h->0) 4/ (sqrt(4(x+h)-5)+sqrt(4x-5)#

#d/dx( sqrt(4x-5)) = 2/ sqrt(4x-5)#