## Condition of the problem

To derive the equation of the normal to the given curve at the point with abscissa $x_{0}$.

$$
y=\frac{1+\sqrt{x}}{1-\sqrt{x}}, x_{\bar{u}}=4
$$
## Solution

Let's find $y^{\prime}:$

$$
\begin{aligned}
& y^{\prime}=\left(\frac{1+\sqrt{x}}{1-\sqrt{x}}\right)^{\prime}=\frac{(1+\sqrt{x})^{\prime}(1-\sqrt{x})-(1+\sqrt{x})(1-\sqrt{x})^{\prime}}{(1-\sqrt{x})^{2}}= \\
& =\frac{\frac{1}{2 \sqrt{x}} \cdot(1-\sqrt{x})-(1+\sqrt{x})\left(-\frac{1}{2 \sqrt{x}}\right)}{(1-\sqrt{x})^{2}}=\frac{1-\sqrt{x}+1+\sqrt{x}}{2 \sqrt{x}(1-\sqrt{x})^{2}}= \\
& =\frac{1}{\sqrt{x}(1-\sqrt{x})^{2}}
\end{aligned}
$$

Then:

$$
y_{0}^{\prime}=y^{\prime}\left(x_{0}\right)=\frac{1}{\sqrt{x_{0}}\left(1-\sqrt{x_{0}}\right)^{2}}=\frac{1}{\sqrt{4}(1-\sqrt{4})^{2}}=\frac{1}{2(1-2)^{2}}=\frac{1}{2}
$$

Since $y^{\prime}\left(x_{\bar{U}}\right) \neq 0$, the equation of the normal line is:

$$
y-y_{\overline{0}}=-\frac{1}{y_{0}^{\prime}} \cdot\left(x-x_{0}\right)
$$

$$
\begin{aligned}
& y_{0}^{\prime}=\frac{1}{2} \\
& y_{0}=y\left(x_{0}\right)=\frac{1+\sqrt{x_{\bar{U}}}}{1-\sqrt{x_{0}}}=\frac{1+\sqrt{4}}{1-\sqrt{4}}=\frac{1+2}{1-2}=-3
\end{aligned}
$$

We get:

$$
\begin{aligned}
& y-(-3)=-\frac{1}{\left(\frac{1}{2}\right)} \cdot(x-4) \\
& y=-2 \cdot(x-4)-3 \\
& y=-2 x+8-3 \\
& y=-2 x+5
\end{aligned}
$$

Thus, the equation of the normal line is:

$y=-2 x+5$

## Problem Kuznetsov Differentiation 3-7