解：
本题属于复合函数求导！
f(x)=1/√(1-x²)
f'(x)= [-1/(1-x²)]*[√(1-x²)]'
=[-1/(1-x²)] * [(1/2)/√(1-x²)] *[1-x²]'
=[-1/(1-x²)] * [(1/2)/√(1-x²)] * -(2x)
=x [(1-x²)^(-3/2)]
本题属于复合函数求导！
f(x)=1/√(1-x²)
f'(x)= [-1/(1-x²)]*[√(1-x²)]'
=[-1/(1-x²)] * [(1/2)/√(1-x²)] *[1-x²]'
=[-1/(1-x²)] * [(1/2)/√(1-x²)] * -(2x)
=x [(1-x²)^(-3/2)]
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第1个回答 2013-03-21
y=1/√(1-x)
=(1-x)^(-1/2)
y'=-1/2*(1-x)^(-1/2-1)*(1-x)'
=-1/2*(1-x)^(-3/2)*(-1)
=(1-x)^(-3/2)/2
=1/[2(1-x)√(1-x)]
=(1-x)^(-1/2)
y'=-1/2*(1-x)^(-1/2-1)*(1-x)'
=-1/2*(1-x)^(-3/2)*(-1)
=(1-x)^(-3/2)/2
=1/[2(1-x)√(1-x)]
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