Question

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

Answer 1

`f'(x) = 4x`

Explanation:

By definition, `f'(x)= lim_(h->0)(f(x+h)-f(x))/h`

so for `f(x) = 2x^2 +1` we have:
`f'(x) = lim_(h->0)( { (2(x+h)^2 +1) - (2x^2 +1)} )/h`
`f'(x) = lim_(h->0)( (2(x^2+2hx+h^2) +1 ) - 2x^2+1)/h`
`f'(x) = lim_(h->0)( 2x^2+4hx+2h^2 +1 - 2x^2-1 )/h`
`f'(x) = lim_(h->0)( 4hx+2h^2 )/h`
`f'(x) = lim_(h->0)( 4x+2h)`
`f'(x) = 4x`