## Problem Statement

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

$y=\frac{x^{16}+9}{1-5 x^{2}}, x_{0}=1$
## Solution

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

$$
\begin{aligned}
& y^{\prime}=\left(\frac{x^{16}+9}{1-5 x^{2}}\right)^{\prime}=\frac{\left(x^{16}+9\right)^{\prime}\left(1-5 x^{2}\right)-\left(x^{16}+9\right)\left(1-5 x^{2}\right)^{\prime}}{\left(1-5 x^{2}\right)^{2}}= \\
& =\frac{16 x^{15}\left(1-5 x^{2}\right)-\left(x^{16}+9\right)(-10 x)}{\left(1-5 x^{2}\right)^{2}}=\frac{16 x^{15}-80 x^{17}+10 x^{17}+90 x}{\left(1-5 x^{2}\right)^{2}}=
\end{aligned}
$$

$=\frac{16 x^{15}-70 x^{17}+90 x}{\left(1-5 x^{2}\right)^{2}}$

Then:

$y_{0}^{\prime}=y^{\prime}\left(x_{0}\right)=\frac{16 \cdot 1^{15}-70 \cdot 1^{17}+90 \cdot 1}{\left(1-5 \cdot 1^{2}\right)^{2}}=\frac{16-70+90}{(1-5)^{2}}=\frac{36}{4^{2}}=\frac{9}{4}$

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

$y_{0}^{\prime}=\frac{9}{4}$

$y_{0}=y\left(x_{0}\right)=\frac{1^{16}+9}{1-5 \cdot 1^{2}}=\frac{1+9}{1-5}=\frac{10}{-4}=-\frac{5}{2}$

We get:

$y-\left(-\frac{5}{2}\right)=\frac{9}{4} \cdot(x-1)$

$y=\frac{9}{4} \cdot x-\frac{9}{4}-\frac{5}{2}$

$y=\frac{9}{4} \cdot x-\frac{19}{4}$

Thus, the equation of the tangent line is:

$y=\frac{9}{4} \cdot x-\frac{19}{4}$

## Problem Kuznetsov Differentiation 3-18