A problem of higher derivatives, solved!

f(x)=arcsinx, which is a McLaughlin series.

Then the coefficient to the power n is multiplied by n! Be what you want.

You haven't used me for a long time.

I'll just add :

f'=1/sqrt(1-x^2)

Again(1-x) a=1-ax+..1)^n*a*(a-1)*.a-n+1)/n!*x^n+..

Equivalent to a -1 2 here

Exchange x 2 for x

f’＝.1)^n*a*(a-1)*.a-n+1)/n!*x^2n+..

Integral. f＝..1)^n*a*(a-1)*.a-n+1)/(2n*n!)*x^(2n+1)+.

Replace n with kf.1)^k*a*(a-1)*.a-k+1)/(2k*k!)*x^(2k+1)+.

From the above, the coefficient to the nth power is as:

When n=2k+1, it is p (-1) k*a*(a-1)*a-k+1)/(2k*k!And what you ask for is n! *p

When n 2k, an even number in the instant, it is obviously 0

Find the first derivative y'＝1/√（1-x^2)

i.e. y'*√（1-x^2)=1

On the left, the (n-1) derivative is found using the Leibniz formula.

y(n)+(n-1)[-x (1-x 2)]+=0y(n) denotes the nth derivative.

Since u= (1-x 2) gives the 1 nth derivative x=0 which is equal to 0, the reason is the presence of the xr factor. So that the left side is 0 at x=0 from the second term

So y(n)=0 and n>=2

As stated above: When n 1, y(n)=1

When n>=2, y(n)=0

Because f(x)=arcsinx

So f'(x)=1/√(1-x^2)

So f"(x)=x/(1-x^2)^(3/2)f"'(x)=(2x^2+1)/(1-x^2)^(5/2)f""(x)=(6x 3+9x) (1-x 2) (7 2) So from the above, we can see:

When n is an even number.

f^n(0)=0

When n is an odd number.

f^n(0)=1

This problem is to find the derivative normally, mainly the combinatorial function, when the derivative of the combinatorial function, just come one by one, divided into three parts, the first, x, the second e to the x power.

The third, ln x, I wrote down the process on paper.

If you don't understand, you can ask questions.

This is the derivative of the product of the three functions, and the specific steps are as follows:

y=xe^ⅹlnx

y′=(xe^x)＇1nⅹ+xe^x*(1/x)(e^x+xe^x)1nx+e^x

e^ⅹ(1+ⅹ)1nx+e^x

e^ⅹ(lnⅹ+ⅹ1nx+1)

This problem uses the derivative of the product of functions, i.e.

uv) = u v+u, where u, denotes two functions.

dy dx= (dy dt) (dx dt), then do the math yourself, this is the formula, you can't be wrong;

d2y d2x answer = d (dy dx) dx where dy dx has been found in the first question, it is an expression about t, but here it is a derivative of x, there is a law for deriving composite functions, [f'(x)=f'(t)*t'(x)] then d(dy dx) dx=[d(dy dx) dt]*(dt dx) is only in parentheses and is very easy to find, as long as dt dx can be found.

It is also known that dx dt=1+2t, and the derivation of the inverse function: [y'(x)*x'(y)=1], we can know that dt dx=1 (1+2t), and then take it step by step.

d 2x dy 2=d(dx dy) dy (find one more derivative for the first derivative) =[d(1 y') dx]*(dx dy) (according to the composite derivative, first the derivative of x and then the derivative of y).

-y''/y'^2)*(1/y')

y''/y'^3。

The topic is in**ah, the topic is in**.

e saw it.

According to the Leibniz formula, you will find that the derivative of the equation about x2 is greater than the third order and it will be 0. Therefore, there are only 3 terms of the higher-order derivative, and when you bring in x=0, there is only one term.

The fourth derivative of 1 gives the coefficient to the 4th power of x.