Example 1. Find the first and second derivatives of the implicit function

$$
\ln \sqrt{x^{2}+y^{2}}-\operatorname{arctg} \frac{y}{x}=0
$$
Solution. In this case, $F(x, y)=\ln \sqrt{x^{2}+y^{2}}-\operatorname{arctg} \frac{y}{x}$. The derivative is defined by formula (7.16). We find the partial derivatives of this function with respect to $x$ and $y$:

$$
\begin{gathered}
F_{x}^{\prime}=\left[\frac{1}{2} \ln \left(x^{2}+y^{2}\right)-\operatorname{arctg} \frac{y}{x}\right]_{x}^{\prime}= \\
=\frac{1 \cdot 2 x}{2\left(x^{2}+y^{2}\right)}-\frac{1}{1+\left(\frac{y}{x}\right)^{2}} \cdot\left(\frac{y}{x^{2}}\right)=\frac{x+y}{x^{2}+y^{2}} \\
F_{y}^{\prime}=\frac{1}{2\left(x^{2}+y^{2}\right)} \cdot 2 y-\frac{1}{1+\left(\frac{y}{x}\right)^{2}} \cdot \frac{1}{x}=\frac{y}{x^{2}+y^{2}}-\frac{x}{x^{2}+y^{2}}=\frac{y-x}{x^{2}+y^{2}} \\
\frac{d y}{d x}=y^{\prime}=-\left(\frac{x+y}{x^{2}+y^{2}}\right):\left(\frac{y-x}{x^{2}+y^{2}}\right)=\frac{x+y}{x-y}
\end{gathered}
$$

We find the second derivative:

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

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