How do you find f'(x) using the definition of a derivative f(x) = x^2 + xf(x)=x2+x?

2 Answers
Jul 23, 2018

2x+12x+1.

Explanation:

Recall that, f'(x)=lim_(t to x){f(t)-f(x)}/(t-x)............(ast).

f(x)=x^2+x rArr f(t)=t^2+t.

:. f(t)-f(x)=(t^2-x^2)+(t-x),

=(t-x)(t+x)+(t-x),

rArr f(t)-f(x)=(t-x){(t+x)+1}.

:. {f(t)-f(x)}/(t-x)={(t+x)+1}, (t!=x).

:.," by "(ast), "f'(x)=lim_(t to x){(t+x)+1},

i.e., f'(x)={(x+x)+1}=2x+1, as desired!

color(blue)("Enjoy Maths.!")

Jul 23, 2018

f'(x)=2x+1

Explanation:

"differentiating from first principles"

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

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

=lim_(hto0)(cancel(x^2)+2hx+h^2cancel(+x)+hcancel(-x^2)cancel(-x))/h

=lim_(hto0)(cancel(h)(2x+h+1))/cancel(h)

=2x+1