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Question #f6a18

1 Answer

Apr 8, 2017

$\frac{d}{d x} (ln \sqrt{x^{2} + 1}) = \frac{x}{x^{2} + 1}$ $d/dx(ln sqrt(x^2+1))=x/(x^2+1)$

Explanation:

By rewriting a square-root as a 1/2-power and ${ln x}^{r} = r ln x$ $ln x^r=r ln x$ ,

$ln \sqrt{x^{2} + 1} = {ln (x^{2} + 1)}^{\frac{1}{2}} = \frac{1}{2} ln (x^{2} + 1)$ $lnsqrt(x^2+1)=ln(x^2+1)^(1/2)=1/2 ln(x^2+1)$

By Log Rule & Chain Rule,

$f'(x)=1/cancel(2)cdot1/(x^2+1)cdot cancel(2)x=x/(x^2+1)$

I hope that this was clear.

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