First Principles Example 3: square root of x

Key Questions

  • How do I find the derivative of f(x) = sqrt(x+3)f(x)=√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