To solve this problem, we need to follow two steps:

Given:

The function is $y = \frac{\sqrt{x}}{x + 5}$.

We are tasked with determining:

1. The equation of the tangent line at the point $\left(9, \frac{3}{14} \right)$.
2. The equation of the normal line at the same point.

Step 1: Tangent Line

The equation of the tangent line at a point $(x_1, y_1)$ is given by: $y - y_1 = m_t(x - x_1)$ Where $m_t$ is the slope of the tangent line, which is found by differentiating the given function.

Differentiate $y = \frac{\sqrt{x}}{x + 5}$

We can apply the quotient rule for differentiation:

$\frac{d}{dx} \left( \frac{\sqrt{x}}{x + 5} \right) = \frac{(x + 5) \cdot \frac{d}{dx}(\sqrt{x}) - \sqrt{x} \cdot \frac{d}{dx}(x+5)}{(x + 5)^2}$

• The derivative of $\sqrt{x}$ is $\frac{1}{2\sqrt{x}}$.
• The derivative of $x + 5$ is 1.

Thus, the derivative (slope of the tangent line) is: $\frac{d}{dx} \left( \frac{\sqrt{x}}{x + 5} \right) = \frac{(x + 5) \cdot \frac{1}{2\sqrt{x}} - \sqrt{x}}{(x + 5)^2}$

Evaluate the derivative at $x = 9$

Substitute $x = 9$ into the derivative to find the slope of the tangent line $m_t$.

$m_t = \frac{(9 + 5) \cdot \frac{1}{2\sqrt{9}} - \sqrt{9}}{(9 + 5)^2}$ $m_t = \frac{14 \cdot \frac{1}{6} - 3}{14^2}$ $m_t = \frac{\frac{14}{6} - 3}{196} = \frac{\frac{7}{3} - 3}{196} = \frac{\frac{7}{3} - \frac{9}{3}}{196} = \frac{-\frac{2}{3}}{196} = -\frac{2}{588} = -\frac{1}{294}$

Thus, the slope of the tangent line is $m_t = -\frac{1}{294}$.

Now, we use the point-slope form of the tangent line equation with the point $\left(9, \frac{3}{14}\right)$:

$y - \frac{3}{14} = -\frac{1}{294}(x - 9)$ Simplify to get the equation of the tangent line.

Step 2: Normal Line

The slope of the normal line is the negative reciprocal of the tangent line slope. Since $m_t = -\frac{1}{294}$, the slope of the normal line $m_n$ is:

$m_n = 294$

Now use the point-slope form for the normal line:

$y - \frac{3}{14} = 294(x - 9)$ Simplify to get the equation of the normal line.

Would you like me to continue simplifying these equations?