y = sin(xy)
y' = (sin(xy))'
y' = cos(xy) * (xy)'
y' = cos(xy) * (xy' + y)
y' = x*cos(xy)*y' + y*cos(xy)
y' = y*cos(xy) / [1 - x*cos(xy)]
tan(x+y) = x
(tan(x+y))' = (x)'
sec²(x+y) * (x+y)' = 1
sec²(x+y) * (1 + y') = 1
1 + y' = cos²(x+y)
y' = -1 + cos²(x+y) = -(1 - cos²(x+y)) = -sin²(x+y)
+ / -檢視主題
由 benice 於 星期六 十一月 05, 2011 8:37 am
[大學]隱微分
由 as88650m 於 星期五 十一月 04, 2011 7:57 am
以隱微分法求dy/dx
1.y=sin(xy)
2.tan(x+y)=x
1.y=sin(xy)
2.tan(x+y)=x