If y = log left[ frac{x + sqrt{x^2 + 25}}{sqr | vedclass

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Question

(B) Given $y = \log \left[ \frac{x + \sqrt{x^2 + 25}}{\sqrt{x^2 + 25} - x} \right]$.Rationalizing the denominator inside the logarithm:$y = \log \left

Vedclass · Accepted Answer

$\frac{2}{\sqrt{x^2 + 25}}$

Vedclass · Answer

(B) Given $y = \log \left[ \frac{x + \sqrt{x^2 + 25}}{\sqrt{x^2 + 25} - x} \right]$.Rationalizing the denominator inside the logarithm:$y = \log \left[ \frac{(x + \sqrt{x^2 + 25})(x + \sqrt{x^2 + 25})}{(\sqrt{x^2 + 25} - x)(\sqrt{x^2 + 25} + x)} \right]$$y = \log \left[ \frac{(x + \sqrt{x^2 + 25})^2}{(x^2 + 25) - x^2} \right]$$y = \log \left[ \frac{(x + \sqrt{x^2 + 25})^2}{25} \right]$Using logarithmic properties:$y = 2 \log (x + \sqrt{x^2 + 25}) - \log 25$Differentiating with respect to $x$:$\frac{dy}{dx} = 2 \cdot \frac{1}{x + \sqrt{x^2 + 25}} \cdot \frac{d}{dx}(x + \sqrt{x^2 + 25}) - 0$$\frac{dy}{dx} = \frac{2}{x + \sqrt{x^2 + 25}} \cdot \left( 1 + \frac{1}{2\sqrt{x^2 + 25}} \cdot 2x \right)$$\frac{dy}{dx} = \frac{2}{x + \sqrt{x^2 + 25}} \cdot \left( \frac{\sqrt{x^2 + 25} + x}{\sqrt{x^2 + 25}} \right)$$\frac{dy}{dx} = \frac{2}{\sqrt{x^2 + 25}}$

If $y = \log \left[ \frac{x + \sqrt{x^2 + 25}}{\sqrt{x^2 + 25} - x} \right]$,then $\frac{dy}{dx} = \dots$

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