Example. Find the derivative of the function

$$
u=x^{2}-\operatorname{arctg}(y+z)
$$

at the point $A(2,1,1)$ in the direction of the point $B(2,4,-3)$.
Solution.

1. Since the function $u=x^{2}-\operatorname{arctg}(y+z)$ is differentiable at the point $A(2,1,1)$, the derivative of this function in any direction $\vec{l}$ exists at this point, which is determined by formula (1).
2. We find the coordinates of the vector $\vec{l}$. In this case,

$$
\vec{l}=\overline{A B}=\{0,3,-4\}
$$

3. We find the unit vector (orth) $\vec{l}_{0}$:

$$
\overrightarrow{l_{0}}=\frac{\vec{l}}{|\vec{l}|}=\frac{\{0,3,-4\}}{\sqrt{0^{2}+3^{2}+(-4)^{2}}}=\left\{0, \frac{3}{5},-\frac{4}{5}\right\}
$$

4. We calculate the partial derivatives of the function $u=x^{2}-\operatorname{arctg}(y+z)$ at the point $A(2,1,1):$

$$
\begin{gathered}
\left.\frac{\partial u}{\partial x}\right|_{(2,1,1)}=\left.2 x\right|_{(2,1,1)}=4,\left.\quad \frac{\partial u}{\partial y}\right|_{(2,1,1)}=-\left.\frac{1}{1+(y+z)^{2}}\right|_{(2,1,1)}=-\frac{1}{5} \\
\left.\frac{\partial u}{\partial z}\right|_{(2,1,1)}=-\left.\frac{1}{1+(y+z)^{2}}\right|_{(2,1,1)}=-\frac{1}{5} .
\end{gathered}
$$

Then

$$
\left.\operatorname{grad} u\right|_{(2,1,1)}=\left\{4,-\frac{1}{5},-\frac{1}{5}\right\}
$$

5. Substituting the obtained values into formula (1) and calculating the scalar product, we get

$$
\left.\frac{\partial u}{\partial l}\right|_{(2,1,1)}=\left(\left.\operatorname{grad} u\right|_{(2,1,1)}, \vec{l}_{0}\right)=4 \cdot 0+\left(-\frac{1}{5}\right) \cdot \frac{3}{5}+\left(-\frac{1}{5}\right) \cdot\left(-\frac{4}{5}\right)=\frac{1}{25}
$$

Answer. $\left.\frac{\partial u}{\partial l}\right|_{(2,1,1)}=\frac{1}{25}$.

Conditions of the Problems. Find the derivative of the function $u(x, y, z)$ at the point $A$ in the direction of the point $B$.

1. $u=x+\ln \left(z^{2}+y^{2}\right)$
$A(2,1,1), \quad B(0,2,0)$.
2. $u=x^{2} y-\sqrt{x y+z^{2}}$,
$A(1,5,-2), \quad B(1,7,-4)$.
3. $u=\sin (x+2 y)+2 \sqrt{x y z}, \quad A\left(\frac{\pi}{2}, \frac{3 \pi}{2}, 3\right), \quad B\left(\frac{\pi}{2}+4, \frac{3 \pi}{2}+3,3\right)$.
4. $u=x^{3}+\sqrt{y^{2}+z^{2}}$,
$A(1,1,0)$,
$B(1,2,-1)$.
5. $u=\sqrt{x y}+\sqrt{9-z^{2}}$,
$A(1,1,0), \quad B(3,3,-1)$.
6. $u=\ln \left(3-x^{2}\right)+x y^{2} z$,
$A(1,3,2)$
$B(0,5,0)$.
7. $u=x^{2} y^{2} z-\ln (z-1)$,
$A(1,1,2)$
$B(6,-5,2 \sqrt{5}+2)$.
8. $u=\ln \left(x^{2}+y^{2}\right)$,
$A(1,-1,2), \quad B(2,-2,3)$.
9. $u=\ln \left(x+\sqrt{z^{2}+y^{2}}\right)$,
$A(1,-3,4), \quad B(-1,-4,5)$.
10. $u=x y-\frac{x}{z}$,
$A(-4,3,-1), \quad B(1,4,-2)$.

Answers. 1. $-\sqrt{6} / 3$. 2. $\sqrt{2} / 12$. 3. 3. 4. $\sqrt{2} / 2$. 5. $2 / 3$. 6. $-11 / 3$. 7. $-4 / 9$. 8. $2 \sqrt{3} / 3$. 9. $-\sqrt{6} / 60$. 10. $20 \sqrt{3} / 9$.

## 6.4. Derivatives of Composite Functions

PROBLEM STATEMENT. Find the derivatives $z_{x}^{\prime}$ and $z_{y}^{\prime}$ of the function $z=z(u, v)$, where $u=u(x, y)$ and $v=v(x, y)$.

SOLUTION PLAN. Since $z$ is a composite function of two variables $x$ and $y$, its derivatives $z_{x}^{\prime}$ and $z_{y}^{\prime}$ are calculated using the formulas

$$
\begin{aligned}
& \frac{\partial z}{\partial x}=\frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial x}+\frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial x} \\
& \frac{\partial z}{\partial y}=\frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial y}+\frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial y}
\end{aligned}
$$

1. We calculate the partial derivatives

$$
\frac{\partial z}{\partial u}, \quad \frac{\partial z}{\partial v}, \quad \frac{\partial u}{\partial x}, \quad \frac{\partial u}{\partial y}, \quad \frac{\partial v}{\partial x}, \quad \frac{\partial v}{\partial y}
$$

2. We substitute the obtained results into formulas (1) and (2) and write the answer.

Remark. Formulas (1) and (2) can be generalized to functions of any number of variables. For example, if the function $f(u, v, w)$ is given, where $u=u(x, y, t), v=v(x, y, t)$ and $w=w(x, y, t)$, then its partial derivatives $f_{x}^{\prime}, f_{y}^{\prime}, f_{t}^{\prime}$ are calculated using the formulas

$$
\begin{aligned}
& \frac{\partial f}{\partial x}=\frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial x}+\frac{\partial f}{\partial v} \cdot \frac{\partial v}{\partial x}+\frac{\partial f}{\partial w} \cdot \frac{\partial w}{\partial x} \\
& \frac{\partial f}{\partial y}=\frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial y}+\frac{\partial f}{\partial v} \cdot \frac{\partial v}{\partial y}+\frac{\partial f}{\partial w} \cdot \frac{\partial w}{\partial y} \\
& \frac{\partial f}{\partial t}=\frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial t}+\frac{\partial f}{\partial v} \cdot \frac{\partial v}{\partial t}+\frac{\partial f}{\partial w} \cdot \frac{\partial w}{\partial t}
\end{aligned}
$$