A.Math微分1條數

y = tan^2(x)
y' = 2tanx/cos^2(x)
y" = [2cos^2(x)/cos^2(x) + 4tanxcosxsinx]/cos^4(x)
= [2 + 4sin^2(x)]/cos^4(x)
= 2[1 + 2sin^2(x)]/cos^4(x)..............(1)
Since y = tan^2(x) = sin^2(x)/cos^2(x) = [1- cos^2(x)]/cos^2(x)
ycos^2(x) = 1 - cos^2(x)
cos^2(x) = 1/(1+y). sub into (1) we get
y" = 2{1 + 2[1 - 1/(1+y)]}/[1/(1+y)]^2.
After simplification, we get
y" = 2(3y+1)(1+y).

我將 dy/dx 以 D 表示 , ^2 = 2次方
1+ (tan x)^2 = (sec x)^2
y = (tan x)^2
Dy = D(tan x)^2
D = 2(tan x)(sec x)^2
D(D) = D[2(tan x)(sec x)^2]
d^2y/dx^2 = [2(sec x)^2][2(sec x)(sec x tan x)]
d^2y/dx^2 = 2[1+ (tan x)^2 ][2(1+ (tan x)^2)tan x)]
d^2y/dx^2 = 2[1+ y ][2+ 2y ^(3/2)]