求下列导数和微分？

Question

在线等！

Answer 1

这里主要是复合函数的求导或者微分得运算。只要把每个函数分析出来，并且求导正确，最后都可以得到正确的结果。

Answer 2

如图所示

Answer 3

(1)
y=arcsin√(1-x^2)
y'
=(1/x).[√(1-x^2)]'
= (1/x).(1/[2√(1-x^2)]) .(1-x^2)'
= (1/x).(1/[2√(1-x^2)]) .(-2x)
=-1/√(1-x^2)
(2)
y=ln(sin√x)
dy
=[1/(sin√x)] d(sin√x)
=[1/(sin√x)] (cos√x) .d√x
= [cot√x/(2√x)] dx
(3)
y=arctan√[(1-x)/(1+x)]
(tany)^2 = (1-x)/(1+x)
= -1 + 2/(1+x)
2tany.(secy)^2 . y' = -2/(1+x)^2
tany.(secy)^2 . y' = -1/(1+x)^2
y'
=-1/{ (1+x)^2.√[(1-x)/(1+x)]. [ (1-x)/(1+x)+1] }
=-1/{ (1+x)^2.√[(1-x)/(1+x)]. [ 2/(1+x)] }
=-(1/2){ 1/√[(1+x)(1-x)] }
=-(1/2)[ 1/√(1-x^2)]
(4)
x=e^t.sint
dx/dt =( 1+cost).e^t
y=e^t.cost
dy/dt =(1-sint).e^t
dy/dx =(dy/dt)/(dx/dt) =(1-sint)/(1+cost)
(5)
x^y=y^x
ylnx=xlny
两边求导
y/x + (lnx).y' = (x/y)y' +lny
y^2 +xy(lnx)y' = x^2.y' +xy(lny)
[xy(lnx) -x^2]y' = xy(lny) -y^2
y' = [xy(lny) -y^2]/[xy(lnx) -x^2]
(6)
y=ln[x+√(1+x^2)]
y'
={ 1/[x+√(1+x^2)] }. [ 1+ x/√(1+x^2)]
=1/√(1+x^2)
y''
=-(1/2)(1+x^2)^(-3/2) . (2x)
=-x/(1+x^2)^(3/2)