What is the derivative of $ln (\sqrt{x^{2} + 1})$ $ln( sqrt(x^2+1))$ ?

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What is the derivative of $ln (\sqrt{x^{2} + 1})$ $ln( sqrt(x^2+1))$ ?

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Answer 1 · 2018-03-04T00:17:10

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

Explanation:

Using the chain rule:

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

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

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

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

We can also note that using the properties of logarithms:

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

and simplify the passages:

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

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What is the derivative of $ln (\sqrt{x^{2} + 1})$ $ln( sqrt(x^2+1))$ ?

1 Answer

Explanation:

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What is the derivative of $ln (\sqrt{x^{2} + 1})$ $ln( sqrt(x^2+1))$ ?