How you you find the derivative #f(x)=x^2# using First Principles?

2 Answers
Apr 1, 2018

#f'(x)=2x#

Explanation:

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

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

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

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

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

#f'(x)=2x#

Apr 1, 2018

#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#