Y = x ^ 2 / √ 1 + x ^ 2 find the second derivative

Y = x ^ 2 / √ 1 + x ^ 2 find the second derivative

1. This problem is the derivation of compound function;
2. The derivative method of compound function is to use chain derivative;
3. The chain derivation method must be used twice in a row
The specific answers are as follows:

Find the n-th derivative of y = 1 / (1 + x)?

y '=-1/(1+x) ²
y ''=1·2/(1+x) ³
y '''=-1·2·3/(1+x)⁴
………………
y^(n)=(-1)^n·n!/ (1+x)^n

Y = (x ^ n-1) / (x-1) n-th derivative

x^n=（x-1+1)^n
=(x-1)^n+n(x-1)^(n-1)+.+n(x-1)^1+1
Then (x ^ n-1) / (x-1) = (x-1) ^ (n-1) + n (x-1) ^ (n-2) +. N (n-1) (x-1) / 2 + n
N-order inverse equals 0

Find the n-order derivative of y = 1 / X (x + 1)

N-order derivative of 1 / X (x + 1), see Annex:
Attachment: n-order derivative.doc

Simplified root sign (cos2x + SiNx quartic) + root sign (cosx quartic cos2x)

Cos2x + SiNx quartic = 1-2 (SiNx) ^ 2 + (SiNx) ^ 4 = (1 - (SiNx) ^ 2) ^ 2 = (SiNx) ^ 4
Cosx quartic - cos2x = (1 - (cosx) ^ 2) ^ 2 = (cosx) ^ 4
Root sign (cos2x + SiNx quartic) + root sign (cosx quartic - cos2x) = (SiNx) ^ 2 + (cosx) ^ 2 = 1

Given SiNx cosx = root 2 / 2, find the third power of SiNx - the value of the third power of cosx

(sin x-cos x)^2=1/2=sin^2 x+cos^2 x+2 sin x*cos x=1+2 sin x*cos x ∴sinx*cosx=-1/4sin^3 x -cos^3 x =(sin x -cos x )(sin^2 x +sin x*cos x +cos^2 x )=(sin x -cos x )(1+sin x*cos x )=（√2/2）(1-1/4)=(3...