## Problem Statement

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

$y=\frac{x^{5}+1}{x^{4}+1}, x_{0}=1$
## Solution

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

$$
\begin{aligned}
& y^{\prime}=\left(\frac{x^{5}+1}{x^{4}+1}\right)^{\prime}=\frac{\left(x^{5}+1\right)^{\prime}\left(x^{4}+1\right)-\left(x^{5}+1\right)\left(x^{4}+1\right)^{\prime}}{\left(x^{4}+1\right)^{2}}= \\
& =\frac{5 x^{4}\left(x^{4}+1\right)-\left(x^{5}+1\right) \cdot 4 x^{3}}{\left(x^{4}+1\right)^{2}}=\frac{5 x^{8}+5 x^{4}-4 x^{8}-4 x^{3}}{\left(x^{4}+1\right)^{2}}= \\
& =\frac{x^{8}+5 x^{4}-4 x^{3}}{\left(x^{4}+1\right)^{2}}
\end{aligned}
$$

Then:

$y_{0}^{\prime}=y^{\prime}\left(x_{0}\right)=\frac{1^{8}+5 \cdot 1^{4}-4 \cdot 1^{3}}{\left(1^{4}+1\right)^{2}}=\frac{1+5-4}{(1+1)^{2}}=\frac{2}{4}=\frac{1}{2}$

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

$y_{0}^{\prime}=\frac{1}{2}$

$y_{0}=y\left(x_{0}\right)=\frac{1^{5}+1}{1^{4}+1}=\frac{1+1}{1+1}=1$

We get:

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

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

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

Thus, the equation of the tangent line is:

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

## Problem Kuznetsov Differentiation 3-17