1. Find the derivative f' of f(x)=x^2. Then find the slope of the curve y=f(x) at x=3 and write an equation for the tangent line. 2. I know that the derivative of x^2 is 2x and I'm guessing that y=x^2, therfore, (3)^2 = 9 which is the value of y. So now the point is (3,9). How do I find the slope? Do I do rise over run which is 9/3 which is 3? 3. I already know the tangent line equation but the slope is getting me confused. I know slope is y=mx+b But what if this is with derivatives?