□投稿者/ X ファミリー(171回)-(2008/05/04(Sun) 14:02:01)
 | これは又複雑ですね。
(間違っていたらごめんなさい。)
(1)
y'={(1/2)√[{1+x^(1/4)}/{1-x^(1/4)}]}
・[-{(1/4)/x^(3/4)}{1+x^(1/4)}-{(1/4)/x^(3/4)}{1-x^(1/4)}]/{1+x^(1/4)}^2
={(1/2)√[{1+x^(1/4)}/{1-x^(1/4)}]}・{-(1/2)/x^(3/4)}/{1+x^(1/4)}^2
={(-1/4)[{1+x^(1/4)}^(3/4)]/[{x^(3/4)}√{1-x^(1/4)}]
(2)
y'=[(cosx)√{(acosx)^2+(bsinx)^2}-(sinx)(-a^2+b^2)(sinxcosx)/√{(acosx)^2+(bsinx)^2}]/{(acosx)^2+(bsinx)^2}
=[(cosx){(acosx)^2+(bsinx)^2}-(sinx)(-a^2+b^2)(sinxcosx)]
/{(acosx)^2+(bsinx)^2}^(3/2)
=(cosx)[{(acosx)^2+(bsinx)^2}+(a^2-b^2)(sinx)^2]
/{(acosx)^2+(bsinx)^2}^(3/2)
={(a^2)cosx}/{(acosx)^2+(bsinx)^2}^(3/2)
|
|