## Condition of the problem

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

$y=14 \sqrt{x}-15 \sqrt[3]{x}+2, x_{0}=1$
## Solution

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

$y^{\prime}=(14 \sqrt{x}-15 \sqrt[3]{x}+2)^{\prime}=\frac{14}{2 \sqrt{x}}-15 \cdot \frac{1}{3} \cdot x^{-\frac{2}{3}}=\frac{7}{\sqrt{x}}-\frac{5}{\sqrt[3]{x^{2}}}$

Then:

$y_{0}^{\prime}=y^{\prime}\left(x_{0}\right)=\frac{7}{\sqrt{1}}-\frac{5}{\sqrt[3]{1^{2}}}=7-5=2$

![](https://cdn.mathpix.com/cropped/2024_05_22_41fcb48922b41c38bc2fg-02.jpg?height=89&width=1776&top_left_y=2056&top_left_x=150)
The form is:

$$
\begin{aligned}
& y-y_{0}=y_{0}^{\prime}\left(x-x_{0}\right), \text { where } \\
& y_{0}^{\prime}=2 \\
& y_{0}=y\left(x_{0}\right)=14 \sqrt{1}-15 \sqrt[3]{1}+2=14-15+2=1
\end{aligned}
$$

We get:

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

$y=2 x-1$

Thus, the equation of the tangent line is:

$y=2 x-1$

Problem Kuznetsov Differentiation 3-26