## Condition of the problem

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

$y=\sqrt{x}-3 \sqrt[3]{x}, x_{0}=64$
## Solution

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

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

Then:

$y_{0}^{\prime}=y^{\prime}\left(x_{0}\right)=\frac{1}{2 \sqrt{x_{0}}}-\frac{1}{\sqrt[3]{x_{0}^{2}}}=\frac{1}{2 \sqrt{64}}-\frac{1}{\sqrt[3]{64^{2}}}=\frac{1}{16}-\frac{1}{16}=0$

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

$x=x_{0}$

$x=64$

Thus, the equation of the normal line is:

$x=64$

## Problem Kuznetsov Differentiation 3-11