Put simply, if the derivative of $y=x^2$ is $0$ when $x$ is $0$, how does that explain all the other values being greater than zero? From what I know, a derivative is the slope of a line tangent to a function. It can also be explained as the slope from a point on a graph to another point that is very, very close to it. So if the slope is $0$ when $x$ is $0$, then wouldn't there be two values (one right before and one right after) that are equal to zero as well? And it could be said that this value approaches zero, so it pretty much is zero. In that case, wouldn't the next very, very small point also be zero, and so on forever? How would the function ever increase in value?
The derivative of $y=x^2$ and other values of x that equal zero
calculusderivatives
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Best Answer
Actually, it's the slope of a line tangent to the graph of a function.
Not really. The derivative of $f$ at $x_0$ is the limit of the slopes of the lines joining $\bigl(x_0,f(x_0)\bigr)$ to points of the form $\bigl(x,f(x)\bigr)$, as $x$ gets closer to $x_0$. And a limit of things which are all different from $0$ may well be $0$.