## Task Condition

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

$y=x^{2}+8 \sqrt{x}-32, x_{\bar{u}}=4$
## Solution

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

$$
y^{\prime}=\left(x^{2}+8 \sqrt{x}-32\right)^{\prime}=2 x+\frac{8}{2 \sqrt{x}}=2 x+\frac{4}{\sqrt{x}}
$$

Then

$$
y_{0}^{\prime}=y^{\prime}\left(x_{0}\right)=2 x_{\overline{0}}+\frac{4}{\sqrt{x_{0}}}=2 \cdot 4+\frac{4}{\sqrt{4}}=8+2=10
$$

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

$$
\begin{aligned}
& y-y_{\bar{u}}=-\frac{1}{y_{\overline{0}}^{\prime}} \cdot\left(x-x_{\bar{u}}\right), \text { where } \\
& y_{\bar{u}}^{\prime}=10 \\
& y_{\overline{0}}=y\left(x_{0}\right)=x_{0}^{2}+8 \sqrt{x_{\overline{0}}}-32=4^{2}+8 \sqrt{4}-32=16+16-32=0
\end{aligned}
$$

We get:

$$
y-0=-\frac{1}{10} \cdot(x-4)
$$

$y=-\frac{x}{10}+\frac{2}{5}$

That is, the equation of the normal line is:

$y=-\frac{x}{10}+\frac{2}{5}$

## Problem Kuznetsov Differentiation 3-4