How do you find the derivative using limits of #f(x)=sqrt(x+1)#?

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Answer 1 · 2017-10-16T13:27:35

#f'(x) = 1/(2sqrt(x+1))#

There are a few different versions of the limit definition of a derivative. For this answer, I'll opt for the following definition:

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

Thus, we continue:

#f'(x) = lim_ {h->0} (sqrt((x+h)+1) - sqrt(x+1))/h#

# = lim_ {h->0} (sqrt((x+h)+1) - sqrt(x+1))/h * (sqrt((x+h)+1) + sqrt(x+1))/(sqrt((x+h)+1) + sqrt(x+1)) #

# = lim_ {h->0} ((sqrt((x+h)+1))^2 - (sqrt(x+1))^2)/(h * (sqrt((x+h)+1) + sqrt(x+1)) #

# = lim_ {h->0} (x+h+1 - (x+1))/(h * (sqrt((x+h)+1) + sqrt(x+1)) #

# = lim_ {h->0} (x+h+1 - x - 1)/(h * (sqrt((x+h)+1) + sqrt(x+1)) #

# = lim_ {h->0} cancel(h)/(cancel(h) * (sqrt((x+h)+1) + sqrt(x+1)) #

# = lim_ {h->0} 1/(sqrt((x+h)+1) + sqrt(x+1)) #

# = 1/(sqrt(x+1) + sqrt(x+1)) = 1/(2sqrt(x+1)) #