## Task Condition

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

$$
y=\frac{-2\left(x^{8}+2\right)}{3\left(x^{4}+1\right)}, x_{0}=1
$$
## Solution

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

$$
\begin{aligned}
& y^{\prime}=\left(\frac{-2\left(x^{8}+2\right)}{3\left(x^{4}+1\right)}\right)^{\prime}=\frac{-2\left(x^{8}+2\right)^{\prime} \cdot 3\left(x^{4}+1\right)-(-2) \cdot\left(x^{8}+2\right) \cdot 3\left(x^{4}+1\right)^{\prime}}{9\left(x^{4}+1\right)^{2}}= \\
& =\frac{-16 x^{7} \cdot 3\left(x^{4}+1\right)+2 \cdot\left(x^{8}+2\right) \cdot 12 x^{3}}{9\left(x^{4}+1\right)^{2}}=\frac{-48 x^{11}-48 x^{7}+24 x^{11}+48 x^{3}}{9\left(x^{4}+1\right)^{2}}= \\
& =\frac{-24 x^{11}-48 x^{7}+48 x^{3}}{9\left(x^{4}+1\right)^{2}}=\frac{-24 x^{3} \cdot\left(x^{8}+2 x^{4}-2\right)}{9\left(x^{4}+1\right)^{2}}= \\
& =\frac{-8 x^{3} \cdot\left(x^{8}+2 x^{4}-2\right)}{3\left(x^{4}+1\right)^{2}}
\end{aligned}
$$

Then:

$y_{0}^{\prime}=y^{\prime}\left(x_{0}\right)=\frac{-8 x_{0}^{3} \cdot\left(x_{0}^{8}+2 x_{0}^{4}-2\right)}{3\left(x_{0}^{4}+1\right)^{2}}=\frac{-8 \cdot 1^{3} \cdot\left(1^{8}+2 \cdot 1^{4}-2\right)}{3\left(1^{4}+1\right)^{2}}=$

$=\frac{-8 \cdot(1+2-2)}{3(1+1)^{2}}=\frac{-8}{3 \cdot 4}=-\frac{2}{3}$

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

$$
\begin{aligned}
& y-y_{0}=y_{0}^{\prime}\left(x-x_{0}\right), \text { where } \\
& y_{0}^{\prime}=-\frac{2}{3} \\
& y_{0}=y\left(x_{0}\right)=\frac{-2\left(x_{0}^{8}+2\right)}{3\left(x_{0}^{4}+1\right)}=\frac{-2\left(1^{8}+2\right)}{3\left(1^{4}+1\right)}=\frac{-2(1+2)}{3(1+1)}=\frac{-6}{6}=-1
\end{aligned}
$$

We get:

$$
\begin{aligned}
& y-(-1)=-\frac{2}{3} \cdot(x-1) \\
& y=-\frac{2}{3} \cdot x+\frac{2}{3}-1 \\
& y=-\frac{2}{3} \cdot x-\frac{1}{3}
\end{aligned}
$$

Thus, the equation of the tangent line is:

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

## Problem Kuznetsov Differentiation 3-16