微分三角函的方程式

Question

Ｙ＝arccos((b+acosx)/(a+bcosx)),0小於等於X小於等於PI,A>B>0

請微分,盡可能附過程謝謝.

Answer 1

1.導函數
設u=(b+acosx)/(a+bcosx),
du/dx=[-asinx(a+bcosx)+bsinx(b+acosx)]/(a+bcosx)²
=-(a²-b²)sinx/(a+bcosx)²
dy/dx=dy/du*du/dx
=-1/√(1-u²)*du/dx
(化簡)1-u²=1-(b+acosx)²/(a+bcosx)²=(a²-b²)sin²x/(a+bcosx)²
故dy/dx=√(a²-b²)/(a+bcosx)

2.微分
dy=dy/dx * dx=√(a²-b²)/(a+bcosx) * dx

註:導函數與微分不同喔!