## Task Condition

Derive the equation of the tangent line to the given curve at the point with abscissa $x_{0}$.
$y=3(\sqrt[3]{x}-2 \sqrt{x}), x_{0}=1$
## Solution

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

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

Then:

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

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

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

We get:

$$
\begin{aligned}
& y-(-3)=-2 \cdot(x-1) \\
& y=-2 x+2-3 \\
& y=-2 x-1
\end{aligned}
$$

Thus, the equation of the tangent line is:

$y=-2 x-1$

Problem Kuznetsov Differentiation 3-19