# 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?

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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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+1. I particularly like this solution - it is very easy to see the cancellation of the RHS term on the last step. – Joe Apr 24 '12 at 22:26
@Jay That is my point on "not rushing things". – Pedro Tamaroff Apr 24 '12 at 22:33

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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$$(x+\sqrt{x^2+1})\sqrt{x^2+1}=x\sqrt{x^2+1}+1+x^2$$

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