<s>[INST] A 'should be simple' derivative Help! I can't seem to prove that $$\Bigl[\log(x+\sqrt{x^2 + 1})\Bigr]' = \frac{1}{\sqrt{x^2 + 1}}$$ I keep getting some horrible answer namely $$\frac{x + \sqrt{x^2 + 1}}{x\sqrt{x^2 + 1} + 1 + x^2}$$ does this cancel down at all?
 [/INST] we have in your denominator
$$ x\sqrt{x^2 + 1} + (1 + x^2) = \bigl(x + \sqrt{x^2 + 1}\bigr)\cdot \sqrt{x^2 + 1}$$
so by canceling $x + \sqrt{x^2 + 1}$ we get $\frac 1{\sqrt{x^2 + 1}}$ as wished.
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