y＝ln[x+√(1+x²)]
∴y'＝[x+√(1+x²)]'/[x+√(1+x²)]
＝[1+x/√(1+x²)]/[x+√(1+x²)]
＝[x+√(1+x²)]/[1+x²+x√(1+x²)]
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