<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] The other answers are quite what's necessary, but note that working tidely you get
$$\frac{d}{{dx}}\log \left( {x + \sqrt {1 + {x^2}} } \right) = \frac{{\frac{d}{{dx}}\left( {x + \sqrt {1 + {x^2}} } \right)}}{{x + \sqrt {1 + {x^2}} }}=$$
$$\frac{{1 + \frac{x}{{\sqrt {1 + {x^2}} }}}}{{x + \sqrt {1 + {x^2}} }}$$
Now, before doing anything else, take ${\sqrt {1 + {x^2}} }$ as a factor in the denominator. You get
$$\frac{{1 + \frac{x}{{\sqrt {1 + {x^2}} }}}}{{x + \sqrt {1 + {x^2}} }} = \frac{1}{{\sqrt {1 + {x^2}} }}\frac{{1 + \frac{x}{{\sqrt {1 + {x^2}} }}}}{{1 + \frac{x}{{\sqrt {1 + {x^2}} }}}} = \frac{1}{{\sqrt {1 + {x^2}} }}$$
Sometimes it is useful not to "rush it" when solving problems or evaluating expressions.
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