a-maths~~~

(1)
y = arccos (1/x)
cos y = 1/x --> sin y = √(x^2 - 1) / x
differentiate both sides w.r.t. x
sin y (dy/dx) = -1/(x^2)
dy/dx = - 1 / (x^2 sin y) = - 1/ {x^2 sin [√(x^2-1)/x)]}
(2)
y =√( a^2 + x^2) / x
dy/dx = {x [2x (0.5)/ √(a^2 + x^2)] - √(a^2 + x^2)} / (x^2)
= (x^2 - a^2 - x^2) / [x^2 √(a^2 + x^2)]
= - a^2 / [x^2 √(a^2 + x^2)]


2008-07-20 08:03:54 補充：
Revised for question 1
(1)

y = arccos (1/x)

cos y = 1/x --> sin y = √(x^2 - 1) / x

differentiate both sides w.r.t. x

- sin y (dy/dx) = -1/(x^2)

dy/dx = 1 / (x^2 sin y) = 1/ {x^2 [√(x^2-1)/x)]} = 1 / x√(x^2 - 1)

樓上那位有不對的地方喔

樓上假設 sin y = [√(x^2 -1) ]/x..........(1)
不過這是錯的!

首先, cos y = x^-1
as (sin y) ^2+ (cos y) ^2= 1
那麼 sin y = √(x^2 -1) / x^2)........(2)
then sin y = √(x^2 -1) /x, 即係(1)

不過,
睇(2), for any values of x, sin y>0

睇(1),when x<0, sin y<0

用呢個contradiction,就可以證明(1)錯左

如果要答案,其實做到 dy/dx = 1/(x^2) sin y 就足夠,
否則就要用absolute sign 囉

2008-07-29 18:43:43 補充：
即係樓上寫既
dy/dx = 1 / x√(x^2 - 1)

再係x度 + absolute, 變成

dy/dx = 1 / |x |√(x^2 - 1)
就ok~

2008-07-29 18:48:40 補充：
至於第2題就冇錯,
其實樓上功力唔差,absolute既問題都諗左我好耐,
因為我之前無諗過拆左個sin y 會走個absolute既問題出黎^^