If y = sqrt{x + sqrt{x^2 + 1}}, then the valu | vedclass

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(A) Given $y = (x + (x^2 + 1)^{1/2})^{1/2}$.Differentiating with respect to $x$ using the chain rule:$\frac{dy}{dx} = \frac{1}{2}(x + (x^2 + 1)^{1/2})

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

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(A) Given $y = (x + (x^2 + 1)^{1/2})^{1/2}$.Differentiating with respect to $x$ using the chain rule:$\frac{dy}{dx} = \frac{1}{2}(x + (x^2 + 1)^{1/2})^{-1/2} \cdot \frac{d}{dx}(x + (x^2 + 1)^{1/2})$$\frac{dy}{dx} = \frac{1}{2y} \cdot (1 + \frac{1}{2}(x^2 + 1)^{-1/2} \cdot 2x)$$\frac{dy}{dx} = \frac{1}{2y} \cdot (1 + \frac{x}{\sqrt{x^2 + 1}})$$\frac{dy}{dx} = \frac{1}{2y} \cdot (\frac{\sqrt{x^2 + 1} + x}{\sqrt{x^2 + 1}})$$\frac{dy}{dx} = \frac{\sqrt{x^2 + 1} + x}{2y\sqrt{x^2 + 1}}$

If $y = \sqrt{x + \sqrt{x^2 + 1}}$, then the value of $\frac{dy}{dx}$ is

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