derivative for parabola

Question

I've been revisiting my calculus and have a simple question I can't seem to answer with respect to derivatives of a parabolic function.

Take: $y=x^2$

Derivative $dy = 2x~dx$

However by simply looking at the graph, if you take $x(1,2) = y(1,4)$, $\dfrac{dy}{dx} = \dfrac{4-1}{2-1} = 3$ and not $4$ ($2x$ at $x = 2$).

Why is the derivative higher at $4$ versus $\dfrac{dy}{dx} = 3$ when looking at the function?

Answer 1

The derivative is the limit when the distance from your point goes to $0$. So $$\frac{dy}{dx}=\lim_{h\to0}\frac{(x+h)^2-x^2}{h}$$ If you choose $h=-1$, you will get a different result than if you have $h=0.1$ or $h=0.01$ or $-0.1$ and $0.01$ on the negative side

Answer 2

3 is the slope of the line connecting $(1,1)$ to $(2,4)$ not the slope of tangent line on parabola in the point $(2,4)$. In fact if we consider the slope of intervening line between $(t,t^2)$ and $(2,4)$ we have$${4-t^2\over2}=2+t$$by tending $t$ to 2 we obtain the slope to be 4 as expected.