How do you use the definition of a derivative to find the derivative of $f(x)= 2x^2 - 3x+4$ ?

Question

1462 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-31T23:57:25

$f'(x) = 4x-3$

By definition $f'(x) =lim_(hrarr0)( (f(x+h)-f(x))/h )$

So, with $f(x)=2x^2-3x+4$ we have:
$f'(x) = lim_(hrarr0)( ( (2(x+h)^2-3(x+h)+4) - (2x^2-3x+4) ) / h )$

$:. f'(x) = lim_(hrarr0)( ( ( 2(x^2+2hx+h^2)-3x-3h+4) - (2x^2-3x+4) ) /h )$

$:. f'(x) = lim_(hrarr0)( ( 2x^2+4hx+2h^2-3x-3h+4 - 2x^2+3x-4) /h )$

$:. f'(x) = lim_(hrarr0)( ( 4hx+2h^2-3h) /h )$

$:. f'(x) = lim_(hrarr0)( 4x+2h-3 )$

$:. f'(x) = 4x-3$

How do you use the definition of a derivative to find the derivative of f(x)= 2x^2 - 3x+4f(x)= 2x^2 - 3x+4?