Find the nth derivative of y arccosx at x 0

Find the derivative first, there is f'(x)=1 (1+x*2), which is f'(x)(1+x*2)=1, then take the n derivatives on both sides, and use the Leibniz formula on the left.

The derivatives of three or more trips with (1+x*2) are zero, so it can be written as f(n+1)(x)(1+x*2)+nf(n)(x)2x+n(n-1)f(n-1)(x)=0, and if you bring 0 into the above equation, you will have f(n+1)(0)=-n(n-1)f(n-1)(0), and then find f(0)=0,f'(0)=1,f"(0)=0, then recursively, and there you have it. Here's the Leibniz formula.

Can't forget.

Finding higher-order derivatives is a major application of Taylor's formula, or power series. The main thing is to take advantage of the uniqueness of expressions. On the one hand, by definition, f(x) arctanx

In McClaulin's formula, the coefficient of x n is: f(n)(0)n! , f(n)(0) denotes the nth derivative at x 0.

On the other hand, fx) 1 (1 x 2) 1) n x (2n), so, f(x) 1) n x (2n 1).

2n 1) Compare the coefficients of x n in the two expressions, and obtain: when n is an even number, the n derivative of f(x) at x 0 is 0; When n is an odd number, let n 2m 1 and the nth derivative of f(x) at x 0 is: (1) m 2m).

A simple calculation is sufficient, and the answer to the head of the limb celery calendar is shown in the first life picture.

There are roughly two ways to do this.

One is by Taylor.

One is to find the nth order directly, and of course you can use some special kind of Silu dispersion, such as sinx cosx in(x+1) and so on.

The first derivative of y (1-x 2) (1 2).

After applying (1+x) a typical.

One more accumulation of points is all you need.

The result of the squared derivative is: 1 (1-x 2)=1 (1-x)*(1+x);

Perform splitting terms: =1 2*(1 1-x.)

1/1+x);

Then believe that you can already see that the problem has been transformed into a request.

1 1-x and.

1 The n-2 derivative of 1+x, this is all regular and formulaic;

For example: [n-2]=(1) n-2

n-2)!1+x) n-1 let x=0, then (-1) n-2n-2)!

There are roughly two ways to do this.

One is by Taylor.

One is to directly seek the nth order.

Of course, with the help of some special formulas.

For example, sinx

cosxin(x+1) and so on.

The first derivative of y.

1-x^2)^(1/2)

Apply (1+x) a again

After the typical formula. Accumulate points one more time.

That's it.

The result of the squared derivative is: 1 (1-x 2)=1 (1-x)*(1+x);

Perform splitting terms: =1 2*(1 1-x.)

1/1+x);

Then believe that you can already see that the problem has been transformed into a request.

1 1-x and 1 1+x

n-2 derivatives, which are all regular and formulaic;

For example: [n-2]=(1) n-2

n-2)!1+x) n-1 let x=0, then (-1) n-2n-2)!

If you have any questions, please feel free to ask.

y'=1/(x^2+1)=1-x^2+x^4-x^6+..1)^nx^(2n)+.

So y'|(x=0)=1

y^(2n)|(x=0)=(1)^n*(2n)!

y^(2n+1)|(x=0)=0

n>=1)

Let's verify it for yourself later).

The derivative of arcsinx is 1 (1-x, and arccosx = 2-arcsinx, then the derivative of arccosx, y'=-1/√(1-x²)。

Let y=arccosx

then cosy=x

Derivative on both sides: siny·y'=1

y'=-1/siny

Since cosy=x, i.e., cosy=x 1=adjacent edge The hypotenuse of the hypotenuse triangle is 1, and the adjacent edge is x, so the opposite side is (1-x) so siny=opposite side hypotenuse = (1-x) 1= (1-x)y'=-1/√(1-x²)

y'=1/(x^2+1)=1-x^2+x^4-x^6+..1)^nx^(2n)+.

So y'|(x=0)=1

y^(2n)|(x=0)=(1)^n*(2n)!

y^(2n+1)|(x=0)=0

n >vertical defect = 1).

After Yu You's defense, verify it and tell the truth).

The derivative of arccosx is: -1 (1-x).

The answer process is as follows:

1) y=arccosx, then cosy=x.

2) Derivative on both sides: -siny·y'=1，y'=-1/siny。

3) Since cosy=x, so siny= modulo positive(1-x)=1-x), so y'=-1/√(1-x²)。

The method is as follows, please comma circle for reference:

If there is help from the landslide, please celebrate.