High math questions, call for help, high math questions, call for help

Tangent equations, right?

f(x) is a continuous function with a period of 5.

f(x)=f(x+5),f(6)=f(1)

f′(6)=lim[f(x+6)-f(6)]/x

x→0lim[f(x+1)-f(1)]/x

x→0f′(1)

f（1+sin x）-3f(1-sin x)=8x+o(x)

When x 0, f(1)-3f(1)=0, then f(1)=limf(x)=0 (obtained from continuous), i.e., f(6)=0

x 0 and f(1+sin x)-3f(1-sin x)=8x+o(x).

f(1+sinx)-f(1)-3f(1-sinx)+3f(1)=8x+o(x)

f(1+sinx)-f(1)]+3f(1)-3f(1-sinx)]=8x+o(x)

lim/sinx=[8x+o(x)]/sinx

x→0f′(1)+3f′(1)=8

f′(1)=2

The tangent equation of the curve y=f(x) at the point (6, f(6)): y-0=2(x-6), i.e., y=2x-12

Since there is f(1+sin x)-3f(1-sin x)=8x+o(x)

limx->0 [f（1+sin x）-3f(1-sin x)]/x=8;

f(x) is derivable at x=1, by Lopida's rule, lim x->0[f'(1+sin x).cos x + 3 f’(1-sin x).cos x]/1 = 8;

Since f(x) is a continuous function, f'(1) +3 f'(1) = 8;

f’（1）=2；f(x) is a continuous function with period 5, so f'(6)=2;

y-f(6)=y-f(1)=2（x-6）；

f(1)-3f(1)=;

The tangent equation for the curve y=f(x) at the points (6, f(6)) is .

y=2x-12;

(1-x^2)^mdx =x(1-x^2)^m|-∫xd(1-x^2)^m

2mx^2(1-x^2)^（m-1)dx=2m∫x^2(1-x^2)^（m-1)dx-2m∫(1-x^2)^(m-1)dx+2m∫(1-x^2)^(m-1)dx

2m(∫(1-x^2)^mdx)+2m∫(1-x^2)^(m-1)dx

Sort it out to get (1-x 2) mdx=2m (2m+1) 1-x 2) (m-1)dx

This results in a recursive and then you can go down and that.

Hello, please add the title.

The study of advanced mathematics has its own characteristics, and the exercises are generally divided into two categories, one is the basic training exercises, which are often attached to each chapter and each section. This kind of problem is relatively simple and not difficult, but it is very important to lay the foundation. The breadth of knowledge is not limited to this chapter and this section, and a variety of mathematical tools are used in solving problems.

The practice of mathematics is an extremely important part of digesting and consolidating knowledge, and it is impossible to achieve the goal without this.

Second, we should pay close attention to the foundation and make gradual progress. In any discipline, the basic content is often the most important part, and it is related to the success or failure of learning. Higher mathematics itself is the foundation of mathematics and other disciplines, and higher mathematics has some important basic content, which is related to the overall situation.

Taking calculus as an example, the limit runs through the whole calculus, the continuity and properties of the function run through a series of theorem conclusions, and the elementary function derivation method and the integration method are related to the future disciplines. Therefore, it is necessary to work hard at the beginning and firmly grasp these basic contents. When learning advanced mathematics, you must take one step at a time, learn and practice in a down-to-earth manner, and the door to success will definitely open to you.

Third, categorize the summary, from thick to thin. The general principle of memorization is to grasp the outline and memorize it in use. Categorization is an important method.

The classification methods of advanced mathematics can be summarized in two parts: content and method, supplemented by examples of representative problems. When categorizing the subsections, special attention should be paid to some conclusions derived from the basic content, that is, the so-called intermediate results, which often appear on some typical examples and exercises, and if you can master some intermediate results, it will be easy to solve general problems and comprehensive training problems.

It is impossible for people to master the knowledge they have learned through a single study, so it is very important to accumulate experience, and the best way is to come to the morning to know the experience!

This problem is mainly based on the method of factorization, where 1-x 3=(1-x)(1+x+x 2), and the specific solution is as follows:

Let's pick C. If you can't type, you can take a look**, I hope it can help you.