First Principles Example 1: x²

Key Questions

  • How do I find the derivative of #x^2 + 7x -4# using first principles?

    First Principles #-># Difference Quotient

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

    #f(x)=x^2+7x-4#

    #f(x+h)=(x+h)^2+7(x+h)-4#

    #f'(x)=lim_(h->0)((x+h)^2+7(x+h)-4-(x^2+7x-4))/h#

    #f'(x)=lim_(h->0)((x+h)^2+7(x+h)-4-x^2-7x+4)/h#

    #f'(x)=lim_(h->0)((x+h)^2+7x+7h-4-x^2-7x+4)/h#

    #f'(x)=lim_(h->0)(x^2+2xh+h^2+7x+7h-4-x^2-7x+4)/h#

    #f'(x)=lim_(h->0)(2xh+h^2+7h)/h#

    #f'(x)=lim_(h->0)(h(2x+h+7))/h#

    #f'(x)=lim_(h->0)(2x+h+7)#

    #f'(x)=2x+(0)+7#

    #f'(x)=2x+7#

    AJ Speller · 1 · Oct 11 2014
  • How you you find the derivative #f(x)=x^2# using First Principles?

    Answer:

    #f'(x)=2x#

    Explanation:

    #f'(x)=lim_(hto0)(f(x+h)-f(x))/h#

    #rArrf'(x)=lim_(hto0)((x+h)^2-x^2)/h#

    #color(white)(rArrf'(x))=lim_(hto0)(cancel(x^2)+2hx+h^2cancel(-x^2))/h#

    #color(white)(rArrf'(x))=lim_(hto0)(cancel(h)(2x+h))/cancel(h)#

    #color(white)(rArrf'(x))=2x#

    Jim G. · 2 · Apr 1 2018

Questions

Derivatives