y=f[(x-2)/(x+2)],f'(x)=arctanx^2,求x=0时y'

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y=f[(x-2)/(x+2)],f'(x)=arctanx^2,求x=0时y'

y=f[(x-2)/(x+2)],f'(x)=arctanx^2,求x=0时y'
y=f[(x-2)/(x+2)],f'(x)=arctanx^2,求x=0时y'

y=f[(x-2)/(x+2)],f'(x)=arctanx^2,求x=0时y'
该函数可看成是复合函数：
y=f(u)
u=u(x)=(x-2)/(x+2)
∴结合题设可知
f'(u)=arctanu²
u'(x)=4/(x+2)²
又当x=0时,u(0)=-1
∴f'(0)
=[arctan(-1)²]×u'(0)
=(arctan1)×1
=arctan1
=π/4
即f'(0)=π/4

y'=f'[(x-2)/(x+2)]*[(x-2)/(x+2)]'
=arctan[(x-2)/(x+2)]²*[1*(x+2)-1*(x-2)]/(x+2)²
所以 x=0
y'=arctan1*4/(0+2)²
=π/4

y=f[(x-2)/(x+2)],=f（x²-4）
y′=f′（x²-4）*2x
=2x*arctan（x²-4）²
x=0 y′=0
希望对你有帮助O(∩_∩)O~

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