First Principles Example 3: square root of x

Key Questions

  • How do I find the derivative of #f(x) = sqrt(x+3)# using first principles?

    Answer:

    #f'(x)=1/(2sqrt(x+3))#

    Explanation:

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

    #f(x)=sqrt(x+3), f(x+h)=sqrt(x+h+3)#, then

    #f'(x)=lim_(h->0)(sqrt(x+h+3)-sqrt(x+3))/h#

    If we evaluate this right away, we get

    #lim_(h->0)(sqrt(x+h+3)-sqrt(x+3))/h=(sqrt(x+3)-sqrt(x+3))/0=0/0#,

    so we need to simplify as this is an indeterminate form.

    Multiply the entire limit by the numerator's conjugate, which is #(sqrt(x+h+3)+sqrt(x+3))/(sqrt(x+h+3)+sqrt(x+3))#. This is the same as multiplying by #1.#

    #f'(x)=lim_(h->0)(sqrt(x+h+3)-sqrt(x+3))/h*(sqrt(x+h+3)+sqrt(x+3))/(sqrt(x+h+3)+sqrt(x+3))#

    The numerator becomes

    #sqrt(x+h+3)-sqrt(x+3) * [sqrt(x+h+3)+sqrt(x+3)]=x+h+3-(x+3)=x+h+3-x-3=h#

    #f'(x)=lim_(h->0)(cancelx+h+cancel3-cancelx-cancel3)/(h(sqrt(x+h+3)+sqrt(x+3))#

    #f'(x)=lim_(h->0)(cancelh)/(cancelh(sqrt(x+h+3)+sqrt(x+3))#

    #f'(x)=lim_(h->0)1/(sqrt(x+h+3)+sqrt(x+3))#

    #f'(x)=1/(sqrt(x+3)+sqrt(x+3))#

    #f'(x)=1/(2sqrt(x+3))#

    VNVDVI · 1 · Apr 20 2018
  • How do I find the derivative of #f(x)=sqrt(x)# using first principles?

    Definition

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

    By Definition,

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

    by multiplying the numerator and the denominator by #sqrt{x+h}+sqrt{x}#,

    #=lim_{h to 0}{sqrt{x+h}-sqrt{x}}/hcdot{sqrt{x+h}+sqrt{x}}/{sqrt{x+h}+sqrt{x}}#

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

    by cancelling out #x#'s and #h#'s,

    #=lim_{h to 0}1/{sqrt{x+h}+sqrt{x}}=1/{sqrt{x+0}+sqrt{x}}=1/{2sqrt{x}}#

    Hence, #f'(x)=1/{2sqrt{x}}#.

    I hope that this was helpful.

    Wataru · 6 · Oct 17 2014

Questions

Derivatives