## Problem Statement

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

$y=6 \sqrt[3]{x}-\frac{16 \sqrt[4]{x}}{3}, x_{0}=1$
## Solution

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

$$
\begin{aligned}
& y^{\prime}=\left(6 \sqrt[3]{x}-\frac{16 \sqrt[4]{x}}{3}\right)^{\prime}=\left(6 \cdot x^{\frac{1}{3}}-\frac{16}{3} \cdot x^{\frac{1}{4}}\right)^{\prime}= \\
& =6 \cdot \frac{1}{3} \cdot x^{-\frac{2}{3}}-\frac{16}{3} \cdot \frac{1}{4} x^{-\frac{3}{4}}=2 \cdot x^{-\frac{2}{3}}-\frac{4}{3} \cdot x^{-\frac{3}{4}}
\end{aligned}
$$

Then:

$y_{0}^{\prime}=y^{\prime}\left(x_{0}\right)=2 \cdot 1^{-\frac{2}{3}}-\frac{4}{3} \cdot 1^{-\frac{3}{4}}=2-\frac{4}{3}=\frac{2}{3}$

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

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

$y_{0}^{\prime}=\frac{2}{3}$

$y_{0}=y\left(x_{0}\right)=6 \sqrt[3]{1}-\frac{16 \sqrt[4]{1}}{3}=6-\frac{16}{3}=\frac{2}{3}$

## We obtain:

$y-\frac{2}{3}=\frac{2}{3} \cdot(x-1)$

$y=\frac{2}{3} \cdot x-\frac{2}{3}+\frac{2}{3}$

$y=\frac{2}{3} \cdot x$

Thus, the equation of the tangent line is:

$y=\frac{2}{3} \cdot x$