Find the equation of the tangent line to the curve $$ y=\sqrt{x}+\frac{1}{\sqrt{x}} $$ at point $$ \left(4, \frac{5}{2}\right) $$ $$ \text { ans: } y=\frac{3}{16} x+\frac{7}{4} $$

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Given the curve
$$
y = \sqrt{x} + \frac{1}{\sqrt{x}} = x^{\frac{1}{2}} + x^{-\frac{1}{2}},
$$
we first find its derivative.

1. **Differentiate the function:**

The derivative of $$ x^{\frac{1}{2}} $$ is
   $$
   \frac{d}{dx}x^{\frac{1}{2}} = \frac{1}{2}x^{-\frac{1}{2}},
   $$
   and the derivative of $$ x^{-\frac{1}{2}} $$ is
   $$
   \frac{d}{dx}x^{-\frac{1}{2}} = -\frac{1}{2}x^{-\frac{3}{2}}.
   $$
   Thus, the derivative of $$ y $$ is
   $$
   y' = \frac{1}{2}x^{-\frac{1}{2}} - \frac{1}{2}x^{-\frac{3}{2}}.
   $$

2. **Evaluate the derivative at $$ x = 4 $$:**

We compute $$ x^{-\frac{1}{2}} $$ and $$ x^{-\frac{3}{2}} $$ at $$ x = 4 $$:
   
   $$
   4^{-\frac{1}{2}} = \frac{1}{\sqrt{4}} = \frac{1}{2},
   $$
   $$
   4^{-\frac{3}{2}} = \frac{1}{4^{\frac{3}{2}}} = \frac{1}{(\sqrt{4})^3} = \frac{1}{8}.
   $$
   
   Substituting into $$ y' $$:
   $$
   y'(4) = \frac{1}{2} \cdot \frac{1}{2} - \frac{1}{2} \cdot \frac{1}{8} = \frac{1}{4} - \frac{1}{16}.
   $$
   
   To combine the fractions, use a common denominator of 16:
   $$
   \frac{1}{4} = \frac{4}{16}, \quad y'(4) = \frac{4}{16} - \frac{1}{16} = \frac{3}{16}.
   $$
   This means the slope $$ m $$ of the tangent line at the point $$ (4, \frac{5}{2}) $$ is
   $$
   m = \frac{3}{16}.
   $$

3. **Write the equation of the tangent line:**

Using the point-slope form
   $$
   y - y_1 = m(x - x_1),
   $$
   where $$ (x_1, y_1) = \left(4, \frac{5}{2}\right) $$ and $$ m = \frac{3}{16} $$, we have:
   $$
   y - \frac{5}{2} = \frac{3}{16}(x - 4).
   $$

4. **Simplify to slope-intercept form:**

Distribute the slope:
   $$
   y - \frac{5}{2} = \frac{3}{16}x - \frac{3}{16} \times 4 = \frac{3}{16}x - \frac{12}{16} = \frac{3}{16}x - \frac{3}{4}.
   $$
   Add $$ \frac{5}{2} $$ to both sides:
   $$
   y = \frac{3}{16}x - \frac{3}{4} + \frac{5}{2}.
   $$
   Express $$ \frac{5}{2} $$ with a denominator of 4:
   $$
   \frac{5}{2} = \frac{10}{4},
   $$
   then combine:
   $$
   -\frac{3}{4} + \frac{10}{4} = \frac{7}{4}.
   $$
   The final equation of the tangent line is:
   $$
   y = \frac{3}{16}x + \frac{7}{4}.
   $$
The equation of the tangent line is $$ y = \frac{3}{16}x + \frac{7}{4} $$.

Find the equation of the tangent line to the curve at point

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