## Problem Statement

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

$y=\frac{1+3 x^{2}}{3+x^{2}}, x_{0}=1$
## Solution

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

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

$=\frac{6 x\left(3+x^{2}\right)-\left(1+3 x^{2}\right) \cdot 2 x}{\left(3+x^{2}\right)^{2}}=\frac{18 x+6 x^{3}-2 x-6 x^{3}}{\left(3+x^{2}\right)^{2}}=$

$=\frac{16 x}{\left(3+x^{2}\right)^{2}}$

Then:

$y_{0}^{\prime}=y^{\prime}\left(x_{0}\right)=\frac{16 x_{0}}{\left(3+x_{0}^{2}\right)^{2}}=\frac{16 \cdot 1}{\left(3+1^{2}\right)^{2}}=\frac{16}{4^{2}}=1$

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

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

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

We get:

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

$y-1=x-1$

$y=x$

Thus, the equation of the tangent line is:

$y=x$

## Problem Kuznetsov Differentiation 3-25