设y=ln(1+a的-2x次方),则f'(0)=

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设y=ln(1+a的-2x次方),则f''(0)=设y=ln(1+a的-2x次方),则f''(0)=设y=ln(1+a的-2x次方),则f''(0)=f''(x)=【1/(1+a的-2x次方)】(a的-2x次

设y=ln(1+a的-2x次方),则f'(0)=
设y=ln(1+a的-2x次方),则f'(0)=

设y=ln(1+a的-2x次方),则f'(0)=
f'(x)=【1/(1+a的-2x次方)】(a的-2x次方)lna ×(-2)
f'(0)=【1/(1+a的0次方)】(a的0次方)lna ×(-2)
=-lna