Question

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

Answer 1

`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`

Answer 2

`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`