设f(x)=tan(x²+1)sec2x,则f´(x)=

设f(x)=tan(x²+1)sec2x,则f´(x)=
设f(x)=tan(x²+1)sec2x,则f´(x)=

设f(x)=tan(x²+1)sec2x,则f´(x)=
f'(x)=[tan(x²+1)]'sec2x+tan(x²+1)(sec2x)'
=sec²(x²+1)*(x²+1)'sec2x+tan(x²+1)sec2xtan2x*(2x)'
=2xsec²(x²+1)sec2x+2tan(x²+1)sec2xtan2x

x=tan(x-y)
x'=tan'(x-y)
1=(x-y)'sec^2(x-y)
1/sec^2(x-y)=1-y'
y'=1-cos^2(x-y)
y'=sin^2(x-y)........................(1)式
y"=sin^2'(x-y)
y"=2sin(x-y)sin'(x-y)
y"=(...

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x=tan(x-y)
x'=tan'(x-y)
1=(x-y)'sec^2(x-y)
1/sec^2(x-y)=1-y'
y'=1-cos^2(x-y)
y'=sin^2(x-y)........................(1)式
y"=sin^2'(x-y)
y"=2sin(x-y)sin'(x-y)
y"=(x-y)'2sin(x-y)cos(x-y)
y"=(1-y')sin2(x-y)
代入(1)式
y"=[1-sin^2(x-y)]sin2(x-y)
y"=cos^2(x-y)sin2(x-y)
y"=sin2(x-y)[cos2(x-y)+1]/2
用万能公式把sin和cos换成tan...........你把这忘了吧呵呵
又因为x=tan(x-y)
所以y"=2x/(1+x^2)[(1-x^2)/(1+x^2)+1]/2
y"=2x/(1+x^2)^2和先反后导一样...............(3)

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