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## 数学代写|微积分代写Calculus代写|Derivatives

In this section we will generalize our results on velocity. This will lead us to the idea of the derivative of a function, which is at the heart of differential calculus.

Go to 147.
147
To launch the discussion, let’s start with a couple of review questions.
When we write $S(t)$ we are stating that position depends on time.
Here position is the dependent variable and time is the variable.
The velocity is the rate of change of position with respect to time. By this we mean that velocity is (give the formal definition again):
$$v(t)=$$
Go to frame 148 for the correct answers.
148
In the last frame you should have written … time is the independent variable, and $v(t)=$ \begin{tabular}{l} $\lim {\Delta t \rightarrow 0} \frac{\Delta S}{\Delta t}$. \ $149 \longrightarrow$ the last frane you should have written …time is the independent variable, and \ \hline \end{tabular} Let us consider any continuous function defined by $\gamma=f(x)$. Here $\gamma$ is our dependent variable, and $x$ is our independent variable. If we ask “At what rate does $\gamma$ change as $x$ changes?,” we can find the answer by taking the following limit: Rate of change of $y$ with respect to $x=\lim {\Delta x \rightarrow 0} \frac{\Delta y}{\Delta x}$.

## 数学代写|微积分代写Calculus代写|Graphs of Functions and Their Derivatives

We have just learned the formal definition of a derivative. Graphically, the derivative of a function $f(x)$ at some value of $x$ is equivalent to the slope of the straight line that is tangent to the graph of the function at that point. Our chief concern in the rest of this chapter will be to find methods for evaluating derivatives of different functions. In doing this it is helpful to have some intuitive idea of how the derivative behaves, and we can obtain this by looking at the graph of the function. If the graph has a steep positive slope, the derivative is large and positive. If the graph has a slight slope downward, the derivative is small and negative. In this section we will get some practice putting to use such qualitative ideas as these, and in the following sections we will learn how to obtain derivatives precisely.

Here is a plot of the simple function $y=x$. We have plotted $\gamma^{\prime}=\frac{d y}{d x}$. Because the slope of $y$ is positive and constant, $\gamma^{\prime}$ is a positive constant.

To prove that $\frac{d}{d x} x=1$, let $\gamma(x)=x$. Then
$$\Delta y=\gamma(x+\Delta x)-\gamma(x)=x+\Delta x-x=\Delta x .$$
Hence,
$$\frac{d y}{d x}=\lim {\Delta x \rightarrow 0} \frac{\Delta y}{\Delta x}=\lim {\Delta x \rightarrow 0} \frac{\Delta x}{\Delta x}=1 .$$
Here is a plot of $y=|x|$. (If you have forgotten the definition of $|x|$, see frame 20.)

Here are sketches of $\gamma=|x|$ and $\gamma^{\prime}$. If you drew this correctly, go on to 164 . If you made a mistake or want further explanation, continue here.As you can see from the graph, $y=|x|=x$ for $x>0$. So for $x>0$ the problem is identical to that in frame 161, and $\gamma^{\prime}=1$. However, for $x<0$, the slope of $|x|$ is negative and is easily seen to be $-1$. At $x=0$, the slope is undefined, for it has the value $+1$ if we approach 0 along the positive $x$-axis and has the value $-1$ if we approach 0 along the negative $x$-axis. Therefore, $\frac{d}{d x}|x|$ is discontinuous at $x=0$. (The function $|x|$ is continuous at this point, but the break in its slope at $x=0$ causes a discontinuity in the derivative.)

# 微积分代考

## 数学代写|微积分代写calculus代写|Derivatives

.

$$v(t)=$$

148

## 数学代写|微积分代写calculus代写|函数及其导数的图

.

$$\Delta y=\gamma(x+\Delta x)-\gamma(x)=x+\Delta x-x=\Delta x .$$

$$\frac{d y}{d x}=\lim {\Delta x \rightarrow 0} \frac{\Delta y}{\Delta x}=\lim {\Delta x \rightarrow 0} \frac{\Delta x}{\Delta x}=1 .$$

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