What are the differences and similarities between finding the derivative using the definition and between finding the derivative using the derivative rules?
What are the differences between the derivative function and a derivative at a point?
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I think I understand what you are asking. As an example, consider the function $f(x)=x^2$. Firstly, we evaluate the derivative by first principles, thus \begin{align} \frac{d}{dx}f(x)&=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\\ &=\lim_{h\to 0}\frac{(x+h)^2-x^2}{h}\\ &=\lim_{x\to 0}\frac{x^2+2xh+h^2-x^2}{h}\\ &=\lim_{h\to 0}(2x+h)\\ &=2x. \end{align} Now, by the standard formula for $f(x)=x^n$, namely $f'(x)=nx^{n-1}:$ \begin{align} \frac{d}{dx}f(x)&=2x^{n-1}=2x^1=2x. \end{align} So, in essence, they are saying precisely the same thing, and the rules of differentiation can be derived (in general) from the standard form of the derivative of a function.
Regarding the difference between the derivative function and the derivative of a point, let us, once again, consider the function $f(x)=x^2$. We just showed that the derivative function $f'$ is given by $$f'(x)=2x.$$ What this tells us is that we can find the value of the derivative at any point of the function (which is just the gradient of the function at that point) $f(x)=x^2$. Hence, if we wanted to find the gradient at the point $x=4$ for the function $f(x)=x^2$, we simply put $x=4$ into the derivative function $f'(x)$, or $$\left.\frac{d}{dx}f(x)\right|_{x=4}=\left.2x\right|_{x=4}=2(4)=8.$$
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Are those the same question?
"What are the differences between the derivative function and a derivative at a point?"
The derivative function is the derivative as defined by differentiating a differentiable function. For example, if f(x) = x^2 + 4, then f'(x) = 2x. So f'(x) is defined over an entire range.
A derivative at a point is simply that: the derivative at a particular point. For example, f'(3.5) = 7
In many situations, a differentiable function is not available, so numerical solutions may be all that is possible. And if many points along a function are required, computing numeric values for derivatives at specific points may be the only way to accomplish the task.
DavidB2013
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