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# Solution for Differentiate Log ( X + √ X 2 + 1 ) ? - CBSE (Commerce) Class 12 - Mathematics

ConceptSimple Problems on Applications of Derivatives

#### Question

Differentiate $\log \left( x + \sqrt{x^2 + 1} \right)$ ?

#### Solution

$\text{Let }y = \log\left( x + \sqrt{x^2 + 1} \right)$

$\text{Differentiate with respect to x we get},$

$\frac{d y}{d x} = \frac{d}{dx}\log\left( x + \sqrt{x^2 + 1} \right)$

$= \frac{1}{x + \sqrt{x^2 + 1}}\frac{d}{dx}\left( x + \left( x^2 + 1 \right)^\frac{1}{2} \right) \left[ \text{Using chain rule} \right]$

$= \frac{1}{x + \sqrt{x^2 + 1}}\left[ 1 + \frac{1}{2} \left( x^2 + 1 \right)^{\frac{1}{2} - 1} \frac{d}{dx}\left( x^2 + 1 \right) \right]$

$= \frac{1}{x + \sqrt{x^2 + 1}}\left[ 1 + \frac{1}{2\sqrt{x^2 + 1}} \times 2x \right]$

$= \frac{1}{x + \sqrt{x^2 + 1}}\left[ \frac{\sqrt{x^2 + 1} + x}{\sqrt{x^2 + 1}} \right]$

$= \frac{1}{\sqrt{x^2 + 1}}$

$So, \frac{d}{dx}\left\{ \log\left( x + \sqrt{x^2 + 1} \right) \right\} = \frac{1}{\sqrt{x^2 + 1}}$

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Solution Differentiate Log ( X + √ X 2 + 1 ) ? Concept: Simple Problems on Applications of Derivatives.
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