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Question

Question
If $y=e^{\sqrt{x}}+e^{-\sqrt{x}}$ then $\frac{d y}{d x}$ equals (A) $\frac{e^{\sqrt{x}}-e^{-\sqrt{x}}}{2 \sqrt{x}}$ (B) $\frac{e^{\sqrt{x}}-e^{-\sqrt{x}}}{2 x}$ (C) $\frac{1}{2 \sqrt{x}} \sqrt{y^{2}-4}$ (D) $\frac{1}{2 \sqrt{x}} \sqrt{y^{2}+4}$

Asked By CosmicGlider20 at

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Felix

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Solution By Steps

Step 1: Differentiate each term

To find

\frac{d y}{d x}, we need to differentiate

y = e^{\sqrt{x}} + e^{-\sqrt{x}} with respect to

x.

Step 2: Apply the chain rule

For

e^{\sqrt{x}}, let

u = \sqrt{x}, then

\frac{d}{dx} e^{\sqrt{x}} = e^{\sqrt{x}} \cdot \frac{d}{dx} \sqrt{x} = e^{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}}.

For

e^{-\sqrt{x}}, let

u = -\sqrt{x}, then

\frac{d}{dx} e^{-\sqrt{x}} = e^{-\sqrt{x}} \cdot \frac{d}{dx} (-\sqrt{x}) = e^{-\sqrt{x}} \cdot \left(-\frac{1}{2\sqrt{x}}\right).

Step 3: Combine the results

Combining the derivatives, we get:

\frac{d y}{d x} = \frac{e^{\sqrt{x}}}{2\sqrt{x}} - \frac{e^{-\sqrt{x}}}{2\sqrt{x}} = \frac{e^{\sqrt{x}} - e^{-\sqrt{x}}}{2\sqrt{x}}

Final Answer

The derivative of

y = e^{\sqrt{x}} + e^{-\sqrt{x}} with respect to

x is:

\boxed{A}

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