Condition of the problem

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

$y=x-x^{3}, x_{0}=-1$
## Solution

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

$y^{\prime}=\left(x-x^{3}\right)^{\prime}=1-3 x^{2}$

Then:

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

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

$y_{0}^{\prime}=-2$

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

We get:

$y-0=-\frac{1}{-2} \cdot(x-(-1))$

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

$y=\frac{x}{2}+\frac{1}{2}$

That is, the equation of the normal line is:

$y=\frac{x}{2}+\frac{1}{2}$

## Kuznetsov Differentiation 3-3