How do you find the derivative of $ln (x + \sqrt{(x^{2}) - 1})$ $ln(x+sqrt((x^2)-1))$ ?

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How do you find the derivative of $ln (x + \sqrt{(x^{2}) - 1})$ $ln(x+sqrt((x^2)-1))$ ?

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Answer 1 · 2016-07-28T01:25:07

$\frac{1}{\sqrt{x^{2} - 1}}$ $1/sqrt(x^2-1)$

Explanation:

$\frac{d}{d x} ln (x + \sqrt{x^{2} - 1}) = \frac{1}{x + \sqrt{x^{2} - 1}} \cdot \frac{d}{d x} (x + \sqrt{x^{2} - 1})$ $d/dx ln(x+sqrt(x^2-1)) = 1/[x+sqrt(x^2-1)] * d/dx(x+sqrt(x^2-1))$ (Standard differential and Chain rule)

$= \frac{1}{x + \sqrt{x^{2} - 1}} \cdot (1 + \frac{1}{2} {(x^{2} - 1)}^{- \frac{1}{2}} \cdot \frac{d}{d x} (x^{2} - 1))$ $= 1/[x+sqrt(x^2-1)] * (1 + 1/2(x^2-1)^(-1/2) * d/dx(x^2-1))$
(Power rule and Chain rule)

$= \frac{1}{x + \sqrt{x^{2} - 1}} \cdot (1 + \frac{1}{2} {(x^{2} - 1)}^{- \frac{1}{2}} \cdot 2 x)$ $= 1/[x+sqrt(x^2-1)] * (1 + 1/2(x^2-1)^(-1/2) * 2x)$

$= 1/[x+sqrt(x^2-1)] (1+ (cancel(2)x)/(cancel(2)sqrt(x^2-1)))$

$= 1/[x+sqrt(x^2-1)] ((sqrt(x^2-1) +x)/sqrt(x^2-1))$

$= 1/[cancel(x+sqrt(x^2-1))] ((cancel(x+ sqrt(x^2-1)))/sqrt(x^2-1))$

$= 1/(sqrt(x^2-1))$

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How do you find the derivative of $ln (x + \sqrt{(x^{2}) - 1})$ $ln(x+sqrt((x^2)-1))$ ?

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Explanation:

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How do you find the derivative of $ln (x + \sqrt{(x^{2}) - 1})$ $ln(x+sqrt((x^2)-1))$ ?