## Condition of the problem

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

$$
y=\frac{x^{29}+6}{x^{4}+1}, x_{0}=1
$$
## Solution

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

$$
\begin{aligned}
& y^{\prime}=\left(\frac{x^{29}+6}{x^{4}+1}\right)^{\prime}=\frac{\left(x^{29}+6\right)^{\prime}\left(x^{4}+1\right)-\left(x^{29}+6\right)\left(x^{4}+1\right)^{\prime}}{\left(x^{4}+1\right)^{2}}= \\
& =\frac{29 x^{28}\left(x^{4}+1\right)-\left(x^{29}+6\right) \cdot 4 x^{3}}{\left(x^{4}+1\right)^{2}}=\frac{29 x^{32}+29 x^{28}-4 x^{32}-24 x^{3}}{\left(x^{4}+1\right)^{2}}= \\
& =\frac{25 x^{32}+29 x^{28}-24 x^{3}}{\left(x^{4}+1\right)^{2}}
\end{aligned}
$$

Then:

$y_{0}^{\prime}=y^{\prime}\left(x_{0}\right)=\frac{25 x_{0}^{32}+29 x_{0}^{28}-24 x_{0}^{3}}{\left(x_{0}^{4}+1\right)^{2}}=\frac{25 \cdot 1^{32}+29 \cdot 1^{28}-24 \cdot 1^{3}}{\left(1^{4}+1\right)^{2}}=$

$=\frac{25+29-24}{(1+1)^{2}}=\frac{30}{4}=7.5$

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

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

$y_{0}^{\prime}=7.5$

$y_{0}=y\left(x_{0}\right)=\frac{x_{0}^{29}+6}{x_{0}^{4}+1}=\frac{1^{29}+6}{1^{4}+1}=\frac{1+6}{1+1}=\frac{7}{2}=3.5$

We get:

$y-3.5=7.5 \cdot(x-1)$

$y-3.5=7.5 x-7.5$

$y=7.5 x-4$

Thus, the equation of the tangent line is:

$y=7.5 x-4$

## Problem Kuznetsov Differentiation 3-14