If y = e^{sqrt{x}} + e^{-sqrt{x}},then frac{d | vedclass

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(D) Given $y = e^{\sqrt{x}} + e^{-\sqrt{x}}$.Differentiating with respect to $x$ using the chain rule:$\frac{dy}{dx} = \frac{d}{dx}(e^{\sqrt{x}}) + \f

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$A$ and $C$

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(D) Given $y = e^{\sqrt{x}} + e^{-\sqrt{x}}$.Differentiating with respect to $x$ using the chain rule:$\frac{dy}{dx} = \frac{d}{dx}(e^{\sqrt{x}}) + \frac{d}{dx}(e^{-\sqrt{x}})$$\frac{dy}{dx} = e^{\sqrt{x}} \cdot \frac{d}{dx}(\sqrt{x}) + e^{-\sqrt{x}} \cdot \frac{d}{dx}(-\sqrt{x})$$\frac{dy}{dx} = e^{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}} + e^{-\sqrt{x}} \cdot (-\frac{1}{2\sqrt{x}})$$\frac{dy}{dx} = \frac{e^{\sqrt{x}} - e^{-\sqrt{x}}}{2\sqrt{x}}$ (This matches option $A$).Now,consider $y^2 - 4 = (e^{\sqrt{x}} + e^{-\sqrt{x}})^2 - 4$$y^2 - 4 = e^{2\sqrt{x}} + e^{-2\sqrt{x}} + 2 - 4 = e^{2\sqrt{x}} + e^{-2\sqrt{x}} - 2$$y^2 - 4 = (e^{\sqrt{x}} - e^{-\sqrt{x}})^2$Taking the square root,$\sqrt{y^2 - 4} = e^{\sqrt{x}} - e^{-\sqrt{x}}$.Substituting this into the derivative expression:$\frac{dy}{dx} = \frac{1}{2\sqrt{x}} \sqrt{y^2 - 4}$ (This matches option $C$).Since both $A$ and $C$ are correct,the correct option is $D$.

$I$. $f'(x) = \frac{3}{4}x^{4/3} + 9$	$II$. $f(x) = \frac{9}{28}x^{7/3} - 2$
$III$. $f(x) = \frac{9}{28}x^{7/3} + 6$	$IV$. $f'(x) = \frac{3}{4}x^{4/3} - 4$

If $y = e^{\sqrt{x}} + e^{-\sqrt{x}}$,then $\frac{dy}{dx}$ equals:

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