Let's take a parabola $y=x^2$. Let's take it's derivative for $x=1$:
We take the derivative of $y=x^2$, and find that as the x-value on the parabola approaches $x=1$ the derivative is 2. $\lim _{h \to 0} = 2$ at point $x=1$ on the parabola.
My question here is why the once we find the derivative, we draw a line with that slope, over that point. Once we find the derivative of $x=1$, we draw a line with gradient 2 over (1,1).
But this makes no sense to me because of 2 reasons:
Isn't the derivative not the slope at a point, but the slope between 2 points as one point gets arbitrarily close to another? So since it fundamentally requires 2 points to do, how can we just draw a line with that slope over 1 point? If there was a slope at the single point, then it would result in the denominator 0, leading to an undefined result!
How can we just assume that the slope at the point is 2, because the derivative as $x+h$ approaches $x$ is 2? Just because the slope is getting closer and closer to 2 as the $x+h$ gets arbitrarily close to $x$ doesn't mean the slope at $x$ is actually going to be 2!
So with all this, can someone explain to me how we can still draw tangent lines that have the slope given by the derivative at a given point. How can we do so when:
The derivative doesn't give the slope at a point, but the slope between 2 points $x$ and $x+h$ when $h$ is arbitrarily small. If there was a slope at the single point, then it would result in the denominator 0, leading to an undefined result!
How can we just assume that the slope at the point is 2, because the derivative as $x+h$ approaches $x$ is 2? Just because the slope is getting closer and closer to 2 as the $x+h$ gets arbitrarily close to x doesn't mean the slope at $x$ is actually going to be 2!
Thank you so much! By the way, can you try to give the explanation without epsilon delta proofs, and just at the level someone learning Khan Academy Calculus?