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

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Answer 1 · 2017-02-04T16:11:53

We use the formula #f'(x) = lim_(h->0) (f(x+ h) - f(x))/h# to find the derivative.

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

Multiply this by the conjugate of the numerator, which is #sqrt(x+ h+ 1) + sqrt(x + 1)#.

#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))/(hsqrt(x + h +1) + hsqrt(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) h/(h(sqrt(x + h + 1) + sqrt(x +1))#

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

You can now evaluate through substitution.

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

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

Hopefully this helps!