use the definition of the derivative to find the derivative f(x)=2x-2

Use the definition of the derivative to find the derivative $$\small{\text{$f(x)=x^2-2$}}$$

$$\\\lim \limits_{h \to 0}
\left[
\frac{f\left( x+h \right) - f\left( x \right)}{h}
\right] \small{\text{$\quad f(x+h) = (x+h)^2-2 \qquad f(x) = x^2-2 $ }}\\\\
\lim \limits_{h \to 0}
\left[
\frac{\left( (x+h)^2-2 \right) - \left( x^2-2 \right)}{h}
\right]
=
\lim \limits_{h \to 0}
\left[
\frac{ x^2+2*x*h+h^2-2 - x^2+2}{h}
\right]
=
\lim \limits_{h \to 0}
\left[
\frac{ 2*x*h+h^2}{h}
\right] \\\\
=
\lim \limits_{h \to 0}
\left[
\frac{ 2*x*h}{h}+\frac{ h^2}{h}
\right]
=
\lim \limits_{h \to 0}
\left[
2*x+ h
\right] = 2x$$

The derivative is the gradient of the tangent to the curve at any given point.

if the curve is a line as it is in f(x)=2x-2 then the gradient stays the same all the time.

The gradient of this line is 2.  Therefore the derivative is 2.  

Melody, the poster seems to have asked one question in the "large print" and a different one in the "fine print"......LOL!!!

Oh yea, I didn't even notice LOL

Well the asker got both questions answered:)