Question

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

Answer 1

`2x+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.!")`

Answer 2

`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`