Example. Find the derivative of the function $y=y(x)$, given implicitly by the equation

$$
\ln \sqrt{x^{2}+y^{2}}=\operatorname{arctg} \frac{y}{x}
$$
SOLUTION.

1. In this case, $F(x, y)=\ln \sqrt{x^{2}+y^{2}}-\operatorname{arctg} \frac{y}{x}$. We compute its partial derivatives:

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

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

It is clear that $F(x, y), F_{x}^{\prime}$, and $F_{y}^{\prime}$ are continuous for all $x \neq 0$ and $F_{y}^{\prime} \neq 0$ when $x \neq y$. Therefore, equation (3) defines a function $y(x)$ that is differentiable at all points $\left(x_{0}, y_{0}\right)$ in the domain where $x \neq 0$ and $x \neq y$.

2. We find $y^{\prime}$ using formula (2)

$$
y^{\prime}=-\frac{F_{x}^{\prime}\left(x_{0}, y_{0}\right)}{F_{y}^{\prime}\left(x_{0}, y_{0}\right)}=-\frac{\left(x_{0}+y_{0}\right) /\left(x_{0}^{2}+y_{0}^{2}\right)}{\left(y_{0}-x_{0}\right) /\left(x_{0}^{2}+y_{0}^{2}\right)}=\frac{x_{0}+y_{0}}{x_{0}-y_{0}}
$$

Answer. $y^{\prime}=\frac{x_{0}+y_{0}}{x_{0}-y_{0}}$ for all $x_{0}, y_{0}$ satisfying equation (3) in the domain where $x \neq 0$ and $x \neq y$.

Conditions of the Problem. Find the derivatives of the functions $y=y(x)$ defined implicitly by the equations.

1. $y^{x}=x^{y}$.
2. $y=1+y^{x}$.
3. $y=x+\ln y$.
4. $x+y=e^{x-y}$.
5. $x^{2} e^{2 y}-y^{2} e^{2 x}=0$.
6. $x-y+\operatorname{arctg} y=0$.
7. $y \sin x-\cos (x-y)=0$.
8. $\sin (x y)-e^{x y}-x^{2} y=0$.
9. $1+x y-\ln \left(e^{x y}+e^{-x y}\right)=0$.
10. $x^{2}-2 x y+y^{2}+x+y-2=0$.

## Answers.

1. $y^{\prime}=-\frac{y^{x} \ln y-y x^{y-1}}{x y^{x-1}-x^{y} \ln x}$.
2. $y^{\prime}=\frac{y^{x} \ln y}{1-x y^{x-1}}$.
3. $y^{\prime}=\frac{y}{y-1}$.
4. $y^{\prime}=\frac{e^{x-y}-1}{e^{x-y}+1}$.
5. $y^{\prime}=\frac{y^{2} e^{2 x}-x e^{2 y}}{x^{2} e^{2 y}-y e^{2 x}}$.
6. $y^{\prime}=\frac{1+y^{2}}{y^{2}}$.
7. $y^{\prime}=\frac{y \cos x+\sin (x-y)}{\sin (x-y)-\sin x}$.
8. $y^{\prime}=\frac{y\left(2 x+e^{x y}-\cos x y\right)}{x\left(\cos x y-e^{x y}-x\right)}$.
9. $y^{\prime}=-\frac{y}{x}$.
10. $y^{\prime}=\frac{2 y-2 x-1}{2 y-2 x+1}$.

## 6.6. Tangent Plane and Normal to a Surface

PROBLEM STATEMENT. Find the equations of the tangent plane and the normal to the surface given by the equation

$$
F(x, y, z)=0
$$

at the point $M\left(x_{0}, y_{0}, z_{0}\right)$.

PLAN OF SOLUTION.

The normal vector to the surface given by the equation

$$
F(x, y, z)=0
$$

at the point $M\left(x_{0}, y_{0}, z_{0}\right)$ is defined by the formula

$$
\vec{n}=\left.\operatorname{grad} F\right|_{M}=\left\{\left.\frac{\partial F}{\partial x}\right|_{M},\left.\frac{\partial F}{\partial y}\right|_{M},\left.\frac{\partial F}{\partial z}\right|_{M}\right\}
$$

Therefore, the equation of the tangent plane to the given surface at the point $M\left(x_{0}, y_{0}, z_{0}\right)$ is

$$
\left.F_{x}^{\prime}\right|_{M}\left(x-x_{0}\right)+\left.F_{y}^{\prime}\right|_{M}\left(x_{0}, y_{0}, z_{0}\right)\left(y-y_{0}\right)+\left.F_{z}^{\prime}\right|_{M}\left(z-z_{0}\right)=0
$$

and the equations of the normal are

$$
\frac{x-x_{0}}{\left.F_{x}^{\prime}\right|_{M}}=\frac{y-y_{0}}{\left.F_{y}^{\prime}\right|_{M}}=\frac{z-z_{0}}{\left.F_{z}^{\prime}\right|_{M}}
$$

1. Find the partial derivatives $F_{x}^{\prime}, F_{y}^{\prime}$, and $F_{z}^{\prime}$ at the point $M\left(x_{0}, y_{0}, z_{0}\right)$.
2. Substitute the found values into equations (1) and (2) and write the answer.

NOTE. If only the values of $x_{0}$ and $y_{0}$ are given, then the coordinate $z_{0}$ of the point $M$ is determined from the condition that the point $M$ belongs to the given surface, i.e., $F\left(x_{0}, y_{0}, z_{0}\right)=0$.