How do you use the definition of a derivative to find the derivative of #f(x) = x + sqrtx#?

1 Answer
Lotusbluete
Feb 4, 2016

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

Explanation:

The definition of a derivative is

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

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

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

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

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

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

... expand the fraction so that you can use the formula #(a+b)(a-b) = a^2 - b^2# afterwards...

# = 1 + lim_(h->0) ((sqrt(x + h) - sqrt(x))* color(blue)((sqrt(x + h) + sqrt(x))))/(h * color(blue)((sqrt(x + h) + sqrt(x))))#

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

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

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

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

At this point, you can safely plug #h = 0# to compute the limit:

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

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

Hope that this helped!

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