## Task Condition

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

$y=-2(\sqrt[3]{x}+3 \sqrt{x}), x_{0}=1$
## Solution

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

$$
\begin{aligned}
& y^{\prime}=(-2(\sqrt[3]{x}+3 \sqrt{x}))^{\prime}=-2\left(x^{\frac{1}{3}}+3 \sqrt{x}\right)^{\prime}= \\
& =-2\left(\frac{1}{3} \cdot x^{-\frac{2}{3}}+\frac{3}{2 \sqrt{x}}\right)=-\frac{2}{3 \sqrt[3]{x^{2}}}-\frac{3}{\sqrt{x}}
\end{aligned}
$$

Then:

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

Since the function $y_{\text {at point }}^{\prime} x_{0}$ has a finite derivative, the equation of the tangent line is:

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

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

We get:

$$
\begin{aligned}
& y-(-8)=-3 \frac{2}{3} \cdot(x-1) \\
& y+8=-\frac{11}{3} \cdot(x-1) \\
& 3 y+24=-11 \cdot(x-1) \\
& 3 y+24=-11 x+11 \\
& 3 y+11 x+13=0
\end{aligned}
$$

That is, the equation of the tangent line is:

$3 y+11 x+13=0$

## Problem Kuznetsov Differentiation 3-24