1. **State the problem:** Find the second derivative $y''$ when $y = e^{\sqrt{x}}$. 2. **Recall the formula:** To find $y''$, we first find $y'$ using the chain rule, then differentiate $y'$ again. 3. **Find the first derivative $y'$:** Given $y = e^{\sqrt{x}}$, let $u = \sqrt{x} = x^{1/2}$. Then $y = e^u$. Using the chain rule: $$y' = e^u \cdot u'$$ Calculate $u'$: $$u' = \frac{d}{dx} x^{1/2} = \frac{1}{2} x^{-1/2} = \frac{1}{2\sqrt{x}}$$ So, $$y' = e^{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}} = \frac{e^{\sqrt{x}}}{2\sqrt{x}}$$ 4. **Find the second derivative $y''$:** Differentiate $y' = \frac{e^{\sqrt{x}}}{2\sqrt{x}}$ using the quotient/product rule. Rewrite $y'$ as: $$y' = \frac{1}{2} e^{\sqrt{x}} x^{-1/2}$$ Use the product rule: $$y'' = \frac{1}{2} \left( \frac{d}{dx} e^{\sqrt{x}} \cdot x^{-1/2} + e^{\sqrt{x}} \cdot \frac{d}{dx} x^{-1/2} \right)$$ Calculate each derivative: - $$\frac{d}{dx} e^{\sqrt{x}} = e^{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}}$$ (chain rule) - $$\frac{d}{dx} x^{-1/2} = -\frac{1}{2} x^{-3/2}$$ Substitute back: $$y'' = \frac{1}{2} \left( e^{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}} \cdot x^{-1/2} + e^{\sqrt{x}} \cdot \left(-\frac{1}{2} x^{-3/2} \right) \right)$$ Simplify powers of $x$: Since $x^{-1/2} = \frac{1}{\sqrt{x}}$, and $\frac{1}{2\sqrt{x}} = \frac{1}{2} x^{-1/2}$, $$y'' = \frac{1}{2} e^{\sqrt{x}} \left( \frac{1}{2} x^{-1/2} \cdot x^{-1/2} - \frac{1}{2} x^{-3/2} \right) = \frac{1}{2} e^{\sqrt{x}} \left( \frac{1}{2} x^{-1} - \frac{1}{2} x^{-3/2} \right)$$ Factor out $\frac{1}{2}$: $$y'' = \frac{1}{2} e^{\sqrt{x}} \cdot \frac{1}{2} \left( x^{-1} - x^{-3/2} \right) = \frac{e^{\sqrt{x}}}{4} \left( x^{-1} - x^{-3/2} \right)$$ 5. **Final answer:** $$\boxed{y'' = \frac{e^{\sqrt{x}}}{4} \left( \frac{1}{x} - \frac{1}{x^{3/2}} \right)}$$ This is the second derivative of $y = e^{\sqrt{x}}$.