If y = frac{x}{2}sqrt{a^2 + x^2} + frac{a^2}{ | vedclass

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(A) Given $y = \frac{x}{2}\sqrt{a^2 + x^2} + \frac{a^2}{2}\log(x + \sqrt{x^2 + a^2})$.Applying the product rule and chain rule to differentiate with r

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

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(A) Given $y = \frac{x}{2}\sqrt{a^2 + x^2} + \frac{a^2}{2}\log(x + \sqrt{x^2 + a^2})$.Applying the product rule and chain rule to differentiate with respect to $x$:$\frac{dy}{dx} = \frac{1}{2} \left[ \sqrt{a^2 + x^2} + x \cdot \frac{1}{2\sqrt{a^2 + x^2}} \cdot 2x \right] + \frac{a^2}{2} \cdot \frac{1}{x + \sqrt{x^2 + a^2}} \cdot \left[ 1 + \frac{1}{2\sqrt{x^2 + a^2}} \cdot 2x \right]$Simplify the first term:$\frac{1}{2} \left[ \frac{a^2 + x^2 + x^2}{\sqrt{a^2 + x^2}} \right] = \frac{a^2 + 2x^2}{2\sqrt{a^2 + x^2}}$Simplify the second term:$\frac{a^2}{2} \cdot \frac{1}{x + \sqrt{x^2 + a^2}} \cdot \left[ \frac{\sqrt{x^2 + a^2} + x}{\sqrt{x^2 + a^2}} \right] = \frac{a^2}{2\sqrt{x^2 + a^2}}$Adding both parts:$\frac{dy}{dx} = \frac{a^2 + 2x^2 + a^2}{2\sqrt{a^2 + x^2}} = \frac{2(a^2 + x^2)}{2\sqrt{a^2 + x^2}} = \sqrt{a^2 + x^2}$.

If $y = \frac{x}{2}\sqrt{a^2 + x^2} + \frac{a^2}{2}\log(x + \sqrt{x^2 + a^2})$,then $\frac{dy}{dx} = $

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