Okay so the problem is: Define the derivative, and demonstrate how you can deduct the formula that gives the derivative of $f(x)=x^2$ , and the rule $(f+g)(x)' = f'(x) + g'(x)$.
I know the derivative is the slope of the tangent line to the graph, but I have no idea how to deduct the formula.
Remember the definition of derivative: $$ f'(x)=\lim_{dx \to 0} \frac{f(x+dx)-f(x)}{dx} $$ Or $$ f'(x)=\lim_{x \to x_0} \frac{f(x)-f(x_0)}{x-x_0} $$
So for $f(x)=x^2$ we have: $$f'(x)=\lim_{dx \to 0} \frac{f(x+dx)-f(x)}{dx}= \lim_{dx \to 0} \frac{(x+dx)^2-x^2}{dx}=\lim_{dx \to 0} \frac{x^2+2xdx+(dx)^2 -x^2}{dx}=\lim_{dx \to 0} 2x+dx=2x$$
Also for the other proof:
Let h(x) = f(x)+g(x)
Then: $$h'(x)=\lim_{dx \to 0}\frac{h(x+dx) -h(x)}{dx}=\lim_{dx \to 0}\frac{f(x+dx)+g(x+dx) -f(x)-g(x)}{dx} = f'(x)=\lim_{dx \to 0} (\frac{f(x+dx)-f(x)}{dx} + \frac{g(x+dx)-g(x)}{dx})=f'(x)+g'(x) $$