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**AP 微积分**

AP课程全称为Advanced Placement，包含英语、数学、艺术人文和科学等34门学科课程，在每年5月开考。旨在为成绩优秀天资聪慧的高中生提供机会,允许他们在高中时期提前提前选修大学水平的课程。修读AP课程能让申请大学时具有很大优势，根据College Board的官方调查，85%被调查的学校表示喜欢申请时有AP成绩的学生。而且进入大学后可换算成大学学分，免修同类型的课程。提早毕业，节省学费和时间。

今天我们来介绍AP微积分(AB卷)涉及到的乘积法则(莱布尼兹法则)。

**Problem ****▉**

**The Derivative of a Product of Two Functions**

In AP Calculus AB and BC, it is essential to find derivatives of functions. If f(x) is a function, then its derivative can be computed as follows:

In practice, most of the time, we do not use this definition to find the derivatives of functions. Instead, we usually depend on shortcuts and rules. We will focus on the product rule. The product rule allows us to find the derivative of the product of two functions.

If u(x) and v(x) are functions; then the product rule says that the derivative of u(x)v(x) is

this formula is very useful to remember and will definitely show up in various AP exam questions. The best way to remember equation (2) is to just do practise problems, however, understanding how this formula came about usually helps to remember it.

**Geometric Arguement**

Let us picture the function u(x) and v(x) as representing the side lengths of a rectangle, and du and dv be a small change in the function u(x) and u(v) respectively. This is shown in the diagram below.

We can see from the diagram that the product of u(x) and v(x) represents the area of the blue rectangle. Now to find the change in that area, we need to consider the new areas represented by both green and yellow sections:

If the changes are infinitesimally small, then the last term will go to zero (think of multiplying two very small numbers) so this last term can be ignored, then for the derivative of this area, we would get the product rule:

this geometric way of looking at the derivatives of the product of two functions can help you remember it. It is good to point out that Leibniz (one of the founders of Calculus) used an approach like this one.

**Algebraic argument**

While the geometric approach is useful, it may not be rigorous enough to establish the product rule. For a more rigorous approach, we can start from the definition of a limit (1) and work our way from there.

Let f(x) = u(x)v(x) then the derivative is given by

Here, we just substituted the function into the definition of the derivative. Looking at the final expression on the right, it seems there is no way to simplify this any further and it does not look like the product rule at all. We need to do some form of algebraic manipulation; we are going to add zero to the numerator.

Recall this trivial fact yet very useful here, A - A = 0, this is true if A is a number or a function.

So inserting expression (4) into the numerator of expression (3) will not change the expression since we are adding zero. We then get,

So far, we have arranged the terms, now on the last expression we have the limit of a sum, this is the same as the sum of the limits. Mathematically this reads

Both of these limits contain a product and we know the limit of a product is the same as the products of the limits, we get

The terms in bold are just the definitions of the derivative, and the terms in black simplify to the respective functions. This fact is true provided the function is continuous, and it will be true if the function is differentiable. So putting it all together, we get

**Example ▉**

Find the derivative of

Solutions:

In AP Calculus AB and BC, it is essential to find derivatives of functions. If f(x) is a function, then its derivative can be computed as follows:

Then applying the product rule (2), we get

Alternative Solution:

Simplify first before carrying out the derivative