Advertisements
Advertisements
Question
Differentiate \[\log \left( x + \sqrt{x^2 + 1} \right)\] ?
Advertisements
Solution
\[\text{Let }y = \log\left( x + \sqrt{x^2 + 1} \right)\]
\[\text{Differentiate with respect to x we get}, \]
\[\frac{d y}{d x} = \frac{d}{dx}\log\left( x + \sqrt{x^2 + 1} \right)\]
\[ = \frac{1}{x + \sqrt{x^2 + 1}}\frac{d}{dx}\left( x + \left( x^2 + 1 \right)^\frac{1}{2} \right) \left[ \text{Using chain rule} \right]\]
\[ = \frac{1}{x + \sqrt{x^2 + 1}}\left[ 1 + \frac{1}{2} \left( x^2 + 1 \right)^{\frac{1}{2} - 1} \frac{d}{dx}\left( x^2 + 1 \right) \right]\]
\[ = \frac{1}{x + \sqrt{x^2 + 1}}\left[ 1 + \frac{1}{2\sqrt{x^2 + 1}} \times 2x \right]\]
\[ = \frac{1}{x + \sqrt{x^2 + 1}}\left[ \frac{\sqrt{x^2 + 1} + x}{\sqrt{x^2 + 1}} \right]\]
\[ = \frac{1}{\sqrt{x^2 + 1}}\]
\[So, \frac{d}{dx}\left\{ \log\left( x + \sqrt{x^2 + 1} \right) \right\} = \frac{1}{\sqrt{x^2 + 1}}\]
