y'=1/[x+â(1+x²)] ·[x+â(1+x²)]'
=1/[x+â(1+x²)] ·[1+x/â(1+x²)]
=1/[x+â(1+x²)] ·[x+â(1+x²)]/[â(1+x²)]
=1/â(1+x²)
y'(â3)=1/2
y''=-1/2Ã(1+x²)^(-3/2)·2x
=-x/â(1+x²)³
=-â3/8
æ以
K=|y''|/(1+y'²)^(3/2)
=(â3/8)/(5/4)^(3/2)
=â3/5â5
=â15/25
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