1. **Problem:** Find $\frac{dy}{dx}$ for the function $y = \sqrt{x} - \frac{1}{\sqrt{x}}$ and simplify the result. 2. **Recall the formulas:** - The derivative of $x^n$ is $nx^{n-1}$. - Rewrite roots as fractional exponents: $\sqrt{x} = x^{\frac{1}{2}}$ and $\frac{1}{\sqrt{x}} = x^{-\frac{1}{2}}$. 3. **Rewrite the function:** $$y = x^{\frac{1}{2}} - x^{-\frac{1}{2}}$$ 4. **Differentiate term-by-term:** $$\frac{dy}{dx} = \frac{1}{2}x^{\frac{1}{2} - 1} - \left(-\frac{1}{2}\right)x^{-\frac{1}{2} - 1}$$ 5. **Simplify the exponents:** $$\frac{dy}{dx} = \frac{1}{2}x^{-\frac{1}{2}} + \frac{1}{2}x^{-\frac{3}{2}}$$ 6. **Rewrite with radicals:** $$\frac{dy}{dx} = \frac{1}{2\sqrt{x}} + \frac{1}{2x^{\frac{3}{2}}} = \frac{1}{2\sqrt{x}} + \frac{1}{2x\sqrt{x}}$$ 7. **Find common denominator $2x\sqrt{x}$ and combine:** $$\frac{dy}{dx} = \frac{x}{2x\sqrt{x}} + \frac{1}{2x\sqrt{x}} = \frac{x + 1}{2x\sqrt{x}}$$ **Final answer:** $$\boxed{\frac{dy}{dx} = \frac{x + 1}{2x\sqrt{x}}}$$