## Task Condition

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

$y=x+\sqrt{x^{3}}, x_{0}=1$
## Solution

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

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

Then:

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

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

$y-y_{0}=-\frac{1}{y_{0}^{\prime}} \cdot\left(x-x_{0}\right)$, where

$y_{0}^{\prime}=\frac{5}{2}$

$y_{0}=y\left(x_{0}\right)=x_{0}+\sqrt{x_{0}^{3}}=1+\sqrt{1^{3}}=2$

We get:

$y-2=-\frac{1}{\left(\frac{5}{2}\right)} \cdot(x-1)$

$y=-\frac{2}{5} \cdot(x-1)+2$

$y=-\frac{2 x}{5}+\frac{2}{5}+2$

$y=-\frac{2 x}{5}+2 \frac{2}{5}$

Thus, the equation of the normal line is:

$y=-\frac{2 x}{5}+2 \frac{2}{5}$

## Problem Kuznetsov Differentiation 3-5