## 数学代写|微积分代写Calculus代写|MTH125

2022年10月12日

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## 数学代写|微积分代写Calculus代写|Differentiation

We have accomplished a great deal in this chapter. In fact, all the really important new ideas involved in differential calculus have been introduced-limits, slopes of curves, and derivatives-and you are equipped in principle to apply these to solve a great variety of problems. However, using the fundamental definition to calculate the derivative in each problem as it comes along would be time-consuming. It would also be a great waste of time because there are numerous rules and tricks for differentiating apparently complicated functions in a few short steps.

You will learn the most important of these rules in the following sections. You will also learn how to differentiate a few functions that occur so often that it is useful to know and remember their derivatives. These include a few of the trigonometric functions, logarithms, and exponentials. The remaining sections cover some special topics, as well as applications of differential calculus to some problems. By the end of this chapter you should be able to use differential calculus for many applications. Well, let’s get going!

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Can you find the derivative of the following simple function?
$$\begin{gathered} \gamma=a \quad(a \text { is a constant). } \ \gamma^{\prime}={1|x| a|0| \text { none of these }} \end{gathered}$$

To find $\gamma^{\prime}$, we go back to the definition $\frac{d y}{d x}=\lim {\Delta x \rightarrow 0} \frac{\Delta y}{\Delta x}$. If $y=a$, $$\frac{\Delta \gamma}{\Delta x}=\frac{\gamma(x+\Delta x)-\gamma(x)}{\Delta x}=\frac{a-a}{\Delta x}=0 .$$ (Remember that the meaning of $\gamma(x+\Delta x)$ is $y$ evaluated at $x+\Delta x$.) $$\lim {\Delta x \rightarrow 0} \frac{\Delta y}{\Delta x}=\lim _{\Delta x \rightarrow 0} 0=0 .$$

## 数学代写|微积分代写Calculus代写|Some Rules for Differentiation

In this section we are going to learn a number of shortcut rules for differentiation without having to go all the way back to the definition of the derivative each time. Some of these rules are derived here; others are derived in Appendix $\mathbf{A}$.

For the rest of this section, we will let $u(x)$ and $v(x)$ stand for any two functions that depend on $x$.

Sum Rule:
Our first rule will let us evaluate the derivative of the sum of $u(x)$ and $v(x)$ in terms of their derivatives. We will derive the rule here. Let
$$\gamma(x)=u(x)+v(x) .$$
Then
\begin{aligned} \frac{d y}{d x} &=\lim {\Delta x \rightarrow 0} \frac{[u(x+\Delta x)+v(x+\Delta x)-u(x)-v(x)]}{\Delta x} \ &=\lim {\Delta x \rightarrow 0} \frac{[u(x+\Delta x)-u(x)]}{\Delta x}+\lim _{\Delta x \rightarrow 0} \frac{[v(x+\Delta x)-v(x)]}{\Delta x} \ &=\frac{d u}{d x}+\frac{d v}{d x} . \end{aligned}
Hence the rule is
$$\frac{d}{d x}(u+v)=\frac{d u}{d x}+\frac{d v}{d x} .$$
If you would like a rigorous justification of the manipulation of the limits in the above proof, see Appendix A2.

# 微积分代考

## 数学代写|微积分代写Calculus代写|Differentiation

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171

$$\begin{gathered} \gamma=a \quad(a \text { is a constant). } \ \gamma^{\prime}={1|x| a|0| \text { none of these }} \end{gathered}$$

## 数学代写|微积分代写Calculus代写|Some Rules for Differentiation

$$\gamma(x)=u(x)+v(x) .$$

\begin{aligned} \frac{d y}{d x} &=\lim {\Delta x \rightarrow 0} \frac{[u(x+\Delta x)+v(x+\Delta x)-u(x)-v(x)]}{\Delta x} \ &=\lim {\Delta x \rightarrow 0} \frac{[u(x+\Delta x)-u(x)]}{\Delta x}+\lim _{\Delta x \rightarrow 0} \frac{[v(x+\Delta x)-v(x)]}{\Delta x} \ &=\frac{d u}{d x}+\frac{d v}{d x} . \end{aligned}

$$\frac{d}{d x}(u+v)=\frac{d u}{d x}+\frac{d v}{d x} .$$

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## MATLAB代写

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