极限的定义，以及对于速度和变化量的理解

Let's think about how we can calculate the derivative at a point for a function $y=f(x)$.

For example, let's sketch an arbitrary graph (in this case, we use $y = x^2$, but it doesn't really matter) for now, and pick two points $A$ and $B$ on the graph:

Let's say we wanted the derivative at the point $A$. Well, the slope at that point would be a line that is tangent to the graph at $A$ (tangent meaning it intersects the graph only at that point). We don't really have any formulas dealing with tangent lines to graphs yet, but we can approach a tangent line by starting with a secant line (passes through two points).

Let's draw the line $\overline{AB}$ and name a bunch of coordinates:

Don't be intimidated by this plot. What we're doing is fixing $A$ on the graph, and then treating $B$ as another point on the graph by increasing the $x$-coordinate of $A$ by a small quantity $\Delta x$.

What's the slope of $\overline{AB}$? That's not that difficult, we use our slope formula: $$\text{Slope} = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{f(x_0 + \Delta x) - f(x_0)}{x_0 + \Delta x - x_0} = \dfrac{f(x_0 + \Delta x) - f(x_0)}{\Delta x}$$

Our original goal was the find the slope of the tangent line at $A$. Think about what we could do to $\Delta x$ to effectively find this line, and then click "Continue" when you think you have an idea.

What happens when $\Delta x$ becomes smaller? $B$ would then become closer and closer to $A$. When $\Delta x$ finally becomes $0$, $B$ coincides with $A$! This (almost) gives us the desired line! However, we can't use a single point to define a line, since we could set the slope to any arbitrary value. Knowing this, what could we possibly do to find the tangent line?

When $B$ gets closer and closer to $A$, the line $\overline{AB}$ becomes closer and closer to our ideal tangent line, but when $B$ coincides with $A$, we don't have a specific line anymore.

Hmmm...closer and closer and reaching a abnormality...we can use a limit! We can take the limit as $\Delta x$ approaches $0$, and then we'll have our tangent line! Our equation is now: $$\lim_{\Delta x \to 0}\dfrac{f(x_0 + \Delta x) - f(x_0)}{\Delta x}$$ And this, we denote as the derivative of $f(x)$ at $A$. If we put $x$ back in for $x_0$ to cover all cases of $x$ where there exists a derivative, we get that $$\text{Derivative of }f\text{ at } x = f'(x) = \dfrac{df}{dx} = \lim_{\Delta x \to 0}\dfrac{f(x + \Delta x) - f(x)}{\Delta x}$$ You do not need to worry about the $df$ notation, that will be covered in the next section.

Now, lets try taking the derivative of $f(x)=x^2$ at $x=1$:

We know that: $$\begin{aligned} \text{Derivative of }x^2\text{ at } (x=1) = f'(1) = &\lim_{\Delta x \to 0}\dfrac{f(1 + \Delta x) - f(1)}{\Delta x} \\ =& \lim_{\Delta x \to 0}\dfrac{(1 + \Delta x)^2 - 1^2}{\Delta x} \\ =& \lim_{\Delta x \to 0}\dfrac{1 + 2\Delta x + \Delta x^2 - 1}{\Delta x} \\ =& \lim_{\Delta x \to 0}\dfrac{2\Delta x + \Delta x^2}{\Delta x} \\ =& \lim_{\Delta x \to 0}2 + \Delta x \\ =& \boxed{2} \end{aligned}$$

There! So the slope of the line tangent to the point $x=1$ of $f(x)=x^2$ is equal to $2$.

Do you think you can try finding the derivative of some functions? Try some or yourself!

IROC- Instantaneous Rate Of change

IROC is The rate of change at a particular moment which is the same as the value of the derivative at a particular point. For a function, the instantaneous rate of change at a point is the same as the slope of the tangent line at that point of a curve.

$Slope =\frac{y_2-y_1}{x2−x1}$

Using differential calculus, we can determine the limit, or the value that Δy/Δx approaches as Δy and Δx get closer to zero; it follows that this limit is the exact slope of the tangent. If y is dependent on x, then it is sufficient to take the limit where only Δx approaches zero. Therefore, the slope of the tangent is the limit of Δy/Δx as Δx approaches zero, or dy/dx. We call this limit the derivative.

$\frac{dy}{dx}=\lim_{\Delta x\to0}\frac{\Delta y}{\Delta x}$

Its value at a point on the function gives us the slope of the tangent at that point. For example, let y=x2. A point on this function is (-2,4). The derivative of this function is dy/dx=2x. So the slope of the line tangent to y at (-2,4) is 2·(-2) = -4. The equation of this tangent line is: y-4=(-4)(x-(-2)) or y = -4x - 4.

$f(x) =\lim_{h \to 0}\frac{f(x + h) - f(x)}{h - x}$

~ a brief description of a derivative ~

A function which gives the slope of the line tangent to a function. The derivative of a function f(x) commonly written $f'(x)$. For example if you have $f(x)=x^4$the exponent will than go to the front of the x as a slope and the exponent will than be substrate by 1 so it will be $f'(x)=4x^3$

To find the instantaneous rate of change you need to find the average rate of change by substituting the interval (x coordinate) into the given equation. Then, once you find your y coordinate you will find the other coordinates by using the first interval and add it with h. For example:

Find the slope of $f(x) =2x^2$ at x=5

$f(x) =2(5)^2$ $f(x) =50$ (5,50)

Then, take interval x=5 and add h: h+5 After that you will then replug the new interval h+5 into the equation. $f(x) =2(h+5) ^2$ $f (x) =2h^2+20h+50$ the second pair of coordinates will be $(h+5,$$2h^2+20h+50$)

You will have to plug in both of your coordinates into the Slope equation above: (5,50) $(h+5,$$2h^2+20h+50$)

$Slope=\frac{2h^2+20h}{h}$ Once you divide you answer will be 2h+20

Practice problems

I'm stealing a good amount of stuff from Brandon Lin's explanation.

Let's try to calculate the derivative of $f(x)=x^2$ at $x=1.$

So it's enough to compute the slope of the green line. But that seems pretty hard, so what do we do?

Let's add another point, $B.$

Ok! Then let's compute the slope of $AB.$

Let's say we move $B$ slightly closer.

Good! So our intuition tells us that as we move "$B$" closer to $A,$ the slope of line $AB$ gets closer to the slope of the green line. Furthermore, when we make $B$ "infinitely close" to $A$ the slope of $AB$ becomes the slope of the green line, which is exactly what we want.

Great! So now we have a game plan: Compute the slope of $AB$ for various points $B$ on the curve, and then take the limit of these slopes as $B$ goes to $A.$

Exactly. By above we have that the $\lim_{x\to 1} \frac{x^2-1}{x-1} = \lim_{x\to 1} x + 1 = 2.$

Great! You now have found the derivative of a nontrivial function completely from scratch! Check out Brandon Lin's explanation for this concept in general. Don't be intimidated by the notation, you know what's going on now!

What is a derivative, you ask? Why, it's the instantaneous rate of change!

Doesn't quite make sense yet? Ok, there are two classic ways of thinking about this:

1. Imagine you are driving a car, and you have a friend outside watching you. Say you start from rest, and speed up slowly. Your friend sees the car change position faster and faster, also known as increasing in speed (or velocity). As the driver, you can see these changes by looking at the speedometer (and the road, of course). The speedometer value as a function of time is a well-known example of the derivative of the car's position with respect to time.

2. From a geometrical standpoint, you can also think of the derivative as the slope of a tangent line of a function. Functions that increase rather quickly would have large derivatives in such a region, whereas functions that don't change much would have derivatives close to 0 at that region.

In fact, we can approximate slopes of a general function $f$ at $x = a$ with the formula:

$$\frac{f(a+h) - f(a)}{h}$$

where h is some small number representing the difference between two chosen x-coordinates. Of course, using limits, we can give a more precise algebraic definition of the derivative:

$$\lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$$

Definition of Derivative

You're driving your car on a highway. Your initial position is 0, and your position as a function of time $t$ is $t^2$. The graph below shows your car's position as a function of time.

Now let's try to find your speed on the highway. We're going to find your speed at time $t = 2$.

At time 2 your position was 4. At time 4 your position was 16.

From the graph of position as a function of time, we can see that the car is speeding up. The car moves a greater distance from time $t = 3$ to $t = 4$ than it does from time $t = 2$ to $t = 3$.

To find the car's speed at time $t = 2$, we want to know how far the car moves in a very small amount of time. We want to find the average speed between $t = 2$ and $t = 2.000001$ (the more 0s we add, the more accurate we get).

Now suppose that the position of your car, as a function of time $t$, is $f(t)$. The derivative of $f$ at $t$ is the velocity of your car at time $t$.

... And that's it! That's the definition of the derivative, and some intuition for why it's useful in real life!

Let $f(x)$ be a function. The derivative of $f(x)$ at $x=a$, denoted $f'(a)$ is

$$f'(a) = \lim_{h\to 0} \frac{f(a+h)-f(a)}{h}$$

provided the limit exists. If the limit does not exist, then $f'(a)$ does not exist.

IRoC and Rates of Change.

The instantaneous rate of change measures the rate of change, or slope, of a curve at a certain instant. Thus, the instantaneous rate of change is given by the derivative. We use the Instantaneous Rate of Change to find the limit. When finding the IRoC we use points with very little distance between them. We use the following formula to do so: [F(x+h)-F(x)]/h 'H' represents the change in the denominator which needs to be as small as possible, with out being 0, in order to find the limit, as it approaches zero.

The derivative is a function which gives the slope of a curve; that is, the slope of the line tangent to a function. The derivative of a function f at a point x is commonly written f '(x).

For example, if $f(x) = x^3$ then $f '(x) = 3x^2$. The slope of the tangent line when $x = 5$ is $f '(x) = 3·5^2 = 75$.

Now let's say that there a car speeding along a straight road and the equation for it's speed is $s(t)= 4*t^2$ and is measured in meters. What is the average rate of change from 2 seconds to 3 seconds?

The average rate of change for this is equal to the total change in position divided by the total change in time:

$\frac{\text{change in s}}{\text{change in t}}$

=$\frac{s(3)-s(2)}{3-2}$

=$\frac{36-12}{1}$

=24 meters per second

But how fast was the car going at 2 seconds? As was mentioned above, the instantaneous rate of change measures the slope of a curve at a certain instant. Thus, the instantaneous rate of change is given by the derivative. In this case, the instantaneous rate is s'(x)=8(x) .

=$f(s)= 4(t)^2$

=$f'(s)= 8(t)$

=$f'(s)= 8(2)$

=$f'(s)=$ 16 meters per second

1. Instantaneous rate of change known as (IRoC) is the equivalent to finding the slope of a tangent line.

1. For example in the image above, when the difference between the 2 points smaller, the average rate of change (ARoC) will become closer to the IRoC.

1. The equation for instantaneous rate of change is:

In this problem x=3

(Video) Finding Instantaneous Rates of Change Using the Definition of Derivatives

by PatrickJMT

Here's a video by patrickJMT explaining how to find the instantaneous rate of change and how to use it to solve problems.

The problem he look at is as follows:

The rate of production of a toy can be determined by the production formula $P(t)=1200t-50t^2$, where $P(t)$ is the number of units produced in a given time and $t$ is the time in hours since the factory shift began. At what time does the rate of toy production equal $600$ units per hour? At what time doe the rate of toy production stop?

They key phrase in this problem is rate of. This clues us in on that we need to use derivatives. So let's find the derivative of the production formula. $$\lim_{h\rightarrow0}\frac{P(t+h)-P(t)}{h}\\ \begin{align} &=\lim_{h\rightarrow0}\frac{1200(t+h)-50(t+h)^2-(1200t-50t^2)}{h}\\ &=\lim_{h\rightarrow0}\frac{1200t+1200h-50t^2-100th-50h^2-1200t+50t^2}{h}\\ &=\lim_{h\rightarrow0}\frac{1200h-100th-50h^2}{h}\\ &=\lim_{h\rightarrow0}\frac{h(1200-100t-50h)}{h}\\ &=\lim_{h\rightarrow0}1200-100t-50h\\ &=1200-100t-50\color{red}{(0)}\\ &\boxed{=1200-100t} \end{align}$$

So now that we got our derivative we can start answering the questions. The first part asks for the time when the rate of production is equal to $600$. $$1200-100t=600\\ 100t=600\\ \boxed{t=6}$$ $$\text{After }6\text{ hours the production will equal }600\text{ units.}$$

The second question asks for the time when the production stops. This means, when the rate of production is zero. $$1200-100t=0\\ 100t=1200\\ \boxed{t=12}$$ $$\text{After }12\text{ hours the production will stop.}$$

In this video he solves another example with instantaneous rate.

The problem says:

A particle moves along a straight line with its position given by the function $p=t^2-6t+8$, where $p$ is in meters and $t$ is in seconds. When does the velocity equal $30$ meters per second? When is the particle at rest?

Since we're given a formula for position, we can take the derivative to determine the velocity. $$\begin{align} p'(t)&=\lim_{h\rightarrow0}\frac{(t+h)^2-6(t+h)+8-(t^2-6t+8)}{h}\\ &=\lim_{h\rightarrow0}\frac{t^2+2th+h^2-6t-6h+8-t^2+6t-8}{h}\\ &=\lim_{h\rightarrow0}\frac{2th+h^2-6h}{h}\\ &=\lim_{h\rightarrow0}\frac{h(2t+h-6)}{h}\\ &=\lim_{h\rightarrow0}2t+h-6\\ &=2t+\color{red}{(0)}-6\\ p'(t)&=\boxed{v(t)=2t-6} \end{align}$$

The first thing the problem asks is when the velocity equals $30$ meters per second. $$2t-6=30\\ 2t=36\\ \boxed{t=18}$$ $$\text{After }18\text{ seconds the particle will be}\\ \text{traveling at }30\text{ meters per second.}$$

The second question the problem asks is when the particle will be at rest. If the particle is at rest, that means its velocity is zero. $$2t-6=0\\ 2t=6\\ \boxed{t=3}$$ $$\text{After }3\text{ seconds the particle will be at rest.}$$