合成関数の微分，逆関数の微分．

$y=\log(x+\sqrt{x^2+1})$
の微分．


(1) 普通に合成関数の微分．
$y' = \frac{(x+\sqrt{x^2+1})'}{x+\sqrt{x^2+1}} \\
= \frac{1+(\sqrt{x^2+1})'}{x+\sqrt{x^2+1}} \\
= \frac{1 + \frac{(x^2+1)'}{2\sqrt{x^2+1}}}{x+\sqrt{x^2+1}} \\
= \frac{1 + \frac{2x}{2\sqrt{x^2+1}}}{x+\sqrt{x^2+1}} \\
= \frac{1 + \frac{x}{\sqrt{x^2+1}}}{x+\sqrt{x^2+1}} \\
= \frac{\sqrt{x^2+1}(1 + \frac{x}{\sqrt{x^2+1}}) }{ \sqrt{x^2+1}(x+\sqrt{x^2+1})} \\
= \frac{\sqrt{x^2+1} + x}{\sqrt{x^2+1}(x+\sqrt{x^2+1})} \\
= \frac{1}{\sqrt{x^2+1}}
$


(2) 逆関数を使って微分．
$y=\log(x+\sqrt{x^2+1})$
を x について解く．log の定義より，
$e^y = x+\sqrt{x^2+1}$
$e^y - x = \sqrt{x^2+1}$
両辺2乗で，ルートをとる．
$(e^y - x)^2 = x^2+1$
$e^{2y} - 2xe^y +x^2 = x^2+1$
$e^{2y} - 2xe^y = 1$
$- 2xe^y = 1 - e^{2y}$
$2xe^y = -1 + e^{2y}$
$2x = (e^{2y}-1) e^{-y} = e^y-e^{-y} $
$x=\frac{e^y -e^{-y} }{2}$
（これは双曲三角関数 sinh）


したがって，
$\frac{dx}{dy} = \frac{e^y -(-e^{-y}) }{2} = \frac{e^y +e^{-y} }{2}$

よって，
$\frac{dy}{dx}
=\frac{1}{(\frac{dx}{dy})}\\
=\frac{1}{(\frac{e^y +e^{-y} }{2})}\\
=\frac{2}{e^y +e^{-y}}\\
=\frac{2e^y}{e^{2y} +1}$

$e^y = x+\sqrt{x^2+1}$
だったので，
$e^{2y}+1 = (e^y)^2+1 = (x+\sqrt{x^2+1})^2+1 \\
= x^2 + 2x\sqrt{x^2+1} + (x^2+1)+1 \\
=2x^2 + 2x\sqrt{x^2+1} +2 \\
=2(x^2 + x\sqrt{x^2+1}+1) \\
=2((x^2+1) + x\sqrt{x^2+1}) \\
=2((\sqrt{x^2+1})^2 + x\sqrt{x^2+1}) \\
=2\sqrt{x^2+1}(\sqrt{x^2+1} + x)$

となり，
$\frac{dy}{dx} = \frac{2e^y}{e^{2y}+1} \\
=\frac{2(x+\sqrt{x^2+1})}{2\sqrt{x^2+1}(\sqrt{x^2+1} + x)} \\
=\frac{1}{\sqrt{x^2+1}}
$

どっちが，簡単というわけではなさそうだが．