## Solution

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

$$
y^{\prime}=(8 \sqrt[1]{x}-70)^{\prime}=\left(8 \cdot x^{\frac{1}{4}}-70\right)^{\prime}=8 \cdot \frac{1}{4} \cdot x^{-\frac{3}{4}}=\frac{2}{\sqrt[4]{x^{3}}}
$$

Then:

$$
y_{\overline{0}}^{\prime}=y^{\prime}\left(x_{\overline{0}}\right)=\frac{2}{\sqrt[2]{x_{\overline{0}}^{3}}}=\frac{2}{\sqrt[4]{16^{3}}}=\frac{2}{8}=\frac{1}{4}
$$

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

$$
\begin{aligned}
& y-y_{\bar{u}}=-\frac{1}{y_{\bar{u}}^{\prime}} \cdot\left(x-x_{\overline{0}}\right), \text { where } \\
& y_{\overline{0}}^{\prime}=\frac{1}{4} \\
& y_{\overline{0}}=y\left(x_{\overline{0}}\right)=8 \sqrt[1]{x_{\overline{0}}}-70=8 \sqrt[1]{16}-70=8 \cdot 2-70=-54
\end{aligned}
$$

We get:

$$
\begin{aligned}
& y-(-54)=-\frac{1}{\left(\frac{1}{4}\right)} \cdot(x-16) \\
& y=-4 \cdot(x-16)-54 \\
& y=-4 x+64-54 \\
& y=-4 x+10
\end{aligned}
$$

Thus, the equation of the normal line is:

$y=-4 x+10$

## Problem Kuznetsov Differentiation 3-8
## Condition of the problem

To compose the equation of the normal to the given curve at the point with abscissa $\tau_{0}$.

$y=8 \sqrt[1]{x}-70, x_{0}=16$