## Task Condition

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

$y=\sqrt[3]{x^{2}}-20, x_{0}=-8$
## Solution

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

$$
y^{\prime}=\left(\sqrt[3]{x^{2}}-20\right)^{\prime}=\left(x^{\frac{2}{3}-20}\right)^{\prime}=\frac{2}{3} \cdot x^{-\frac{1}{3}}=\frac{2}{3 \sqrt[3]{x}}
$$

Then:

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

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

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

$y_{0}^{\prime}=-\frac{1}{3}$

$y_{0}=y\left(x_{0}\right)=\sqrt[3]{x_{0}^{2}}-20=\sqrt[3]{(-8)^{2}}-20=4-20=-16$

We get:

$$
\begin{aligned}
& y-(-16)=-\frac{1}{\left(-\frac{1}{3}\right)} \cdot(x-(-8)) \\
& y=3 \cdot(x+8)-16 \\
& y=3 x+24-16 \\
& y=3 x+8
\end{aligned}
$$

Thus, the equation of the normal line is:

$y=3 x+8$

## Problem Kuznetsov Differentiation 3-6