Question

How do you find the derivative of `f(x) = x^2+3x+1` using the limit definition?

Answer 1

Do some substituting and algebra to get `f'(x)=2x+3`.

Explanation:

The definition of the derivative is:
`f'(x)=lim_(h->0)(f(x+h)-f(x))/h`

Our function `f(x)` equals `x^2+3x+1`. Applying it to the definition, we have:
`f'(x)=lim_(h->0)((x+h)^2+3(x+h)+1-(x^2+3x+1))/h`

And doing some algebra to finish off:
`f'(x)=lim_(h->0)(x^2+2xh+h^2+3x+3h+1-x^2-3x-1)/h`
`f'(x)=lim_(h->0)(2xh+h^2+3h)/h`
`f'(x)=lim_(h->0)(h(2x+h+3))/h`
`f'(x)=lim_(h->0)2x+h+3`
`f'(x)=2x+(0)+3`
`f'(x)=2x+3`