Hi I'm still struggling with finding the definition of the derivative (from first principles),
find f'(x) of f(x)=-2x^2+4x-7
Given $f(x), f'(x) = \displaystyle{\lim_{h\to 0}}\dfrac{f(x+h)-f(x)}{h}$
$f(x)=-2x^2+4 x - 7$
$f'(x)=\displaystyle{\lim_{h \to 0}}\dfrac{\left(-2(x+h)^2+4(x+h)-7\right)-\left(-2x^2+4x-7\right)}{h}$
$f'(x)=\displaystyle{\lim_{h \to 0}}\dfrac{-4hx +4h-2h^2}{h}$
$f'(x)=\displaystyle{\lim_{h \to 0}}~-4x+4-2h=-4x+4$