Using the limit definition, how do you find the derivative of #f(x) =sqrt (x+1) #?

1 Answer
Dec 25, 2015

To find the derivative, apply the definition:

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

Explanation:

With #f(x) = sqrt(x+1)#, we apply the definition:

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

To handle the terms in the numerator, it looks like we will need to multiply by the conjugate:

#f'(x) =#
#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))#

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

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

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

At this point, we can directly apply the limit to arrive at an answer:

#f'(x) = 1/(sqrt(x + 0 + 1) + sqrt(x + 1))#

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

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

If you are familiar with the chain rule for derivatives, we can use it to test our result:

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