frac{d}{dx} left( log left( sqrt{x + sqrt{x^2 | vedclass

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Question

(C) Let $y = \log \left( \sqrt{x + \sqrt{x^2 + a^2}} \right)$.Using the property of logarithms,$\log(u^n) = n \log(u)$,we can simplify the expression:

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let $y = \log \left( \sqrt{x + \sqrt{x^2 + a^2}} \right)$.Using the property of logarithms,$\log(u^n) = n \log(u)$,we can simplify the expression:$y = \frac{1}{2} \log \left( x + \sqrt{x^2 + a^2} \right)$.Now,differentiate with respect to $x$:$\frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{x + \sqrt{x^2 + a^2}} \cdot \frac{d}{dx} \left( x + \sqrt{x^2 + a^2} \right)$.Calculate the derivative of the inner term:$\frac{d}{dx} \left( x + \sqrt{x^2 + a^2} \right) = 1 + \frac{1}{2 \sqrt{x^2 + a^2}} \cdot (2x) = 1 + \frac{x}{\sqrt{x^2 + a^2}} = \frac{\sqrt{x^2 + a^2} + x}{\sqrt{x^2 + a^2}}$.Substitute this back into the derivative expression:$\frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{x + \sqrt{x^2 + a^2}} \cdot \left( \frac{\sqrt{x^2 + a^2} + x}{\sqrt{x^2 + a^2}} \right)$.Cancel the common term $(x + \sqrt{x^2 + a^2})$:$\frac{dy}{dx} = \frac{1}{2 \sqrt{x^2 + a^2}}$.

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

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