Partial derivatives, differential, directional derivative, gradient.

References:

1. Efimov A.V. (Ed.), Demidovich B.P. (Ed.). Collection of problems in mathematics for higher education institutions. Linear algebra and fundamentals of mathematical analysis. Part 1. 1991. 481 p.

2. B.P. Demidovich. Collection of problems and exercises in mathematical analysis 624 pp. Moscow: "CheRo", 1997

Let $(x_1, ..., x_k, .., x_n)$ be an arbitrary fixed point from the domain of definition of the function $u=f(x_1, ..., x_n)$. Giving the variable $x_k,,, (k=1, 2, ..., n)$ the increment $\delta x_k$, consider the limit. $$\lim\limits_{\delta x_k\rightarrow 0}\frac{f(x_1,...,x_k+\delta x_k,...,x_n)-f(x_1,...,x_k,...,x_n)}{\delta x_k}.$$

This limit is called the partial derivative (of the 1st order) of the given function with respect to the variable $x_k$ at the point $(x_1, ..., x_n)$ and is denoted by $\frac{\partial u}{\partial x_k}$ or $f'_{x_k}(x_1, ..., x_n)$.

Partial derivatives are calculated according to the usual rules and formulas of differentiation (with all variables except $x_k$ treated as constants).

The partial derivatives of the 2nd order of the function $u=f(x_1, x_2, ..., x_n)$ are called the partial derivatives of its partial derivatives of the 1st order. Second-order derivatives are denoted as follows: $$\frac{\partial}{\partial x_k}\left(\frac{\partial u}{\partial x_k}\right)=\frac{\partial^2u}{\partial x^2_k}=f''_{x_kx_k}(x_1, x_2, ..., x_k, ..., x_n).$$ $$\frac{\partial}{\partial x_l}\left(\frac{\partial u}{\partial x_k}\right)=\frac{\partial^2u}{\partial x_k\partial x_l}=f''_{x_kx_l}(x_1, x_2, ..., x_k, ..., x_l, ..., x_n).$$ Similarly, partial derivatives of orders higher than the second are defined and denoted.

Examples:

Find the first and second-order partial derivatives of the given functions:

7.55. $z=x^5+y^5-5x^3y^3.$

Solution.

$$z'_x=(x^5+y^5-5x^3y^3)'_x=5x^4-15x^2y^3;$$

$$z'_y=(x^5+y^5-5x^3y^3)'_y=5y^4-15x^3y^2;$$

$$z'_{xx}=(5x^4-15x^2y^3)'_x=20x^3-30xy^3;$$

$$z'_{xy}=(5x^4-15x^2y^3)'_y=-45x^2y^2;$$

$$z'_{yy}=(5y^4-15x^3y^2)'_y=20y^3-30x^3y;$$

$$z'_{yx}=(5y^4-15x^3y^2)'_x=-45x^2y^2.$$

Answer: $z'_x=5x^4-15x^2y^3;$ $z'_y=5y^4-15x^3y^2;$ $z'_{xx}=20x^3-30xy^3;$ $z'_{xy}=-45x^2y^2;$ $z'_{yy}=20y^3-30x^3y;$ $z'_{yx}=-45x^2y^2.$

7.57. $z=\frac{xy}{\sqrt{x^2+y^2}}.$

Solution.

$$z'_x=\left(\frac{xy}{\sqrt{x^2+y^2}}\right)'_x=\frac{y\sqrt{x^2+y^2}-xy\frac{2x}{2\sqrt{x^2+y^2}}}{x^2+y^2}=\frac{y(x^2+y^2)-x^2y}{(x^2+y^2)\sqrt{x^2+y^2}}=$$ $$=\frac{y^3}{(x^2+y^2)\sqrt{x^2+y^2}};$$

$$z'_y=\left(\frac{xy}{\sqrt{x^2+y^2}}\right)'_y=\frac{x\sqrt{x^2+y^2}-xy\frac{2y}{2\sqrt{x^2+y^2}}}{x^2+y^2}=\frac{x(x^2+y^2)-xy^2}{(x^2+y^2)\sqrt{x^2+y^2}}=$$ $$=\frac{x^3}{(x^2+y^2)\sqrt{x^2+y^2}};$$

$$z'_{xx}=\left(\frac{y^3}{(x^2+y^2)^{3/2}}\right)'_x=-\frac{3}{2}y^32x(x^2+y^2)^{-5/2}=-3y^3x(x^2+y^2)^{-5/2};$$

$$z'_{xy}=\left(\frac{y^3}{(x^2+y^2)^{3/2}}\right)'_y=\frac{3y^2(x^2+y^2)^{3/2}-\frac{3}{2}y^3(x^2+y^2)^{1/2}\cdot 2y}{(x^2+y^2)^3}=$$ $$=\frac{3y^2(x^2+y^2)-3y^4}{(x^2+y^2)^{5/2}}=\frac{3y^2x^2}{(x^2+y^2)^{5/2}};$$

$$z'_{yy}=\left(\frac{x^3}{(x^2+y^2)^{3/2}}\right)'_y=-\frac{3}{2}x^32y(x^2+y^2)^{-5/2}=-3x^3y(x^2+y^2)^{-5/2};$$

$$z'_{yx}=\left(\frac{x^3}{(x^2+y^2)^{3/2}}\right)'_x=\frac{3x^2(x^2+y^2)^{3/2}-\frac{3}{2}x^3(x^2+y^2)^{1/2}\cdot 2x}{(x^2+y^2)^3}=$$ $$=\frac{3x^2(x^2+y^2)-3x^4}{(x^2+y^2)^{5/2}}=\frac{3x^2y^2}{(x^2+y^2)^{5/2}}.$$

Answer: $z'_x=\frac{y^3}{(x^2+y^2)\sqrt{x^2+y^2}};$ $z'_y=\frac{x^3}{(x^2+y^2)\sqrt{x^2+y^2}};$ $z'_{xx}=-3y^3x(x^2+y^2)^{-5/2};$ $z'_{xy}=\frac{3y^2x^2}{(x^2+y^2)^{5/2}};$ $z'_{yy}=-3x^3y(x^2+y^2)^{-5/2};$ $z'_{yx}=\frac{3x^2y^2}{(x^2+y^2)^{5/2}}.$

7.61.$z=\ln(x^2+y^2).$

Solution.

$$z'_x=\left(\ln(x^2+y^2)\right)'_x=\frac{2x}{x^2+y^2};$$

$$z'_y=\left(\ln(x^2+y^2)\right)'_y=\frac{2y}{x^2+y^2};$$

$$z'_{xx}=\left(\frac{2x}{x^2+y^2}\right)'_x=\frac{2(x^2+y^2)-2x\cdot 2x}{x^2+y^2}=\frac{2(-x^2+y^2)}{x^2+y^2};$$

$$z'_{xy}=\left(\frac{2x}{x^2+y^2}\right)'_y=2x\cdot 2y\frac{-1}{(x^2+y^2)^2}=\frac{-4xy}{(x^2+y^2)^2};$$

$$z'_{yy}=\left(\frac{2y}{x^2+y^2}\right)'_x=\frac{2(x^2+y^2)-2y\cdot 2y}{x^2+y^2}=\frac{2(x^2-y^2)}{x^2+y^2};$$

$$z'_{yx}=\left(\frac{2y}{x^2+y^2}\right)'_x=2y\cdot 2x\frac{-1}{(x^2+y^2)^2}=\frac{-4xy}{(x^2+y^2)^2}.$$

Answer: $z'_x=\frac{2x}{x^2+y^2};$ $z'_y=\frac{2y}{x^2+y^2};$ $z'_{xx}=\frac{2(-x^2+y^2)}{x^2+y^2};$ $z'_{xy}=\frac{-4xy}{(x^2+y^2)^2};$ $z'_{yy}=\frac{2(x^2-y^2)}{x^2+y^2};$ $z'_{yx}=\frac{-4xy}{(x^2+y^2)^{2}}.$

7.66. Find $f'_x(3, 2),\, f'_y(3, 2),$ $f'_{xx}(3, 2),\, f'_{xy}(3, 2),$ $f'_{yy}(3, 2),$ if $f(x, y)=x^3y+xy^2-2x+3y-1.$

Solution.

Let's find the partial derivatives:

$$f'_x=\left(x^3y+xy^2-2x+3y-1\right)'_x=3x^2y+y^2;$$

$$f'_y=\left(x^3y+xy^2-2x+3y-1\right)'_y=x^3+2xy+3;$$

$$f'_{xx}=\left(3x^2y+y^2\right)'_x=6xy;$$

$$f'_{xy}=\left(3x^2y+y^2\right)'_y=3x^2+2y;$$

$$f'_{yy}=\left(x^3+2xy+3\right)'_x=2x.$$

Now let's find the values of the partial derivatives at the point $(3, 2):$

$$f'_x(3, 2)=(3x^2y+y^2)|_{(3,2)}=54+4=58;$$

$$f'_y(3, 2)=(x^3+2xy+3)|_{(3,2)}=27+12+3=42;$$

$$f'_{xx}(3, 2)=6xy|_{(3,2)}=36;$$

$$f'_{xy}(3, 2)=(3x^2+2y)|_{(3,2)}=27+4=31;$$

$$f'_{yy}(3, 2)=2x|_{(3,2)}=6.$$

Answer: $f'_x(3, 2)=58,\, f'_y(3, 2)=42,$ $f'_{xx}(3, 2)=36,\, f'_{xy}(3, 2)=31,$ $f'_{yy}(3, 2)=4.$

7.79. Show that $\left(\frac{\partial z}{\partial x}\right)^2+\frac{\partial z}{\partial y}+x+z=0$ given $z=4e^{-2y}+(2x+4y-3)e^{-y}-x-1$.

Solution.

Let's find the partial derivatives:

$$\frac{\partial z}{\partial x}=(4e^{-2y}+(2x+4y-3)e^{-y}-x-1)'_x=2e^{-y}-1;$$

$$\frac{\partial z}{\partial y}=(4e^{-2y}+(2x+4y-3)e^{-y}-x-1)'_y=-8e^{-2y}+4e^{-y}-(2x+4y-3)e^{-y}.$$

$$\left(\frac{\partial z}{\partial x}\right)^2+\frac{\partial z}{\partial y}+x+z=$$ $$=\left(2e^{-y}-1\right)^2+\left(-8e^{-2y}+4e^{-y}-(2x+4y-3)e^{-y}\right)+$$ $$+x+4e^{-2y}+(2x+4y-3)e^{-y}-x-1=$$ $$=4e^{-2y}-4e^{-y}+1-8e^{-2y}+4e^{-y}-2xe^{-y}-4ye^{-y}+3e^{-y}+$$ $$+4e^{-2y}+2xe^{-y}+4ye^{-y}-3e^{-y}-1=0.$$

Answer: Proven.

For the differential of the function $u=f(x_1, x_2,...,x_n)$, the formula applies: $$du=\frac{\partial u}{\partial x_1}dx_1+\frac{\partial u}{\partial x_2}dx_2+...+\frac{\partial u}{\partial x_n}dx_n.$$

Functions $u$ and $v$ of several variables obey the usual rules of differentiation:$$d(u+v)=du+dv,$$ $$d(uv)=vdu+udv,$$ $$d\left(\frac{u}{v}\right)=\frac{vdu-udv}{v^2}.$$

For a sufficiently small $\rho=\sqrt{\Delta x_1^2+\Delta x_2^2+...+\Delta x_n^2}$, for a differentiable function $u=f(x_1, x_2, ..., x_n)$, approximate equalities occur: $$\Delta u\approx du,$$ $$f(x_1+\Delta x_1,\, x_2+\Delta x_2,...,\, x_n+\Delta x_n)\approx f(x_1, x_2, ..., x_n)+df(x_1, x_2, ..., x_n).$$

Examples:

Find differentials of functions:

7.89. $z=\ln(y+\sqrt{x^2+y^2}).$

Solution.

$$dz=z'_xdx+z'_ydy.$$

$$z'_x=(\ln(y+\sqrt{x^2+y^2}))'_x=\frac{1}{y+\sqrt{x^2+y^2}}(y+\sqrt{x^2+y^2})'_x=\frac{1}{y+\sqrt{x^2+y^2}}\frac{1}{2\sqrt{x^2+y^2}}(x^2+y^2)'_x=$$ $$=\frac{1}{y+\sqrt{x^2+y^2}}\frac{1}{2\sqrt{x^2+y^2}}2x=\frac{x}{y\sqrt{x^2+y^2}+x^2+y^2}.$$

$$z'_y=(\ln(y+\sqrt{x^2+y^2}))'_y=\frac{1}{y+\sqrt{x^2+y^2}}(y+\sqrt{x^2+y^2})'_y=$$ $$=\frac{1}{y+\sqrt{x^2+y^2}}\left(1+\frac{1}{2\sqrt{x^2+y^2}}(x^2+y^2)'_y\right)=$$ $$=\frac{1}{y+\sqrt{x^2+y^2}}\left(1+\frac{1}{2\sqrt{x^2+y^2}}2y\right)=\frac{1}{y+\sqrt{x^2+y^2}}+\frac{y}{\sqrt{x^2+y^2}(y+\sqrt{x^2+y^2})}=$$ $$=\frac{\sqrt{x^2+y^2}+y}{(y+\sqrt{x^2+y^2})\sqrt{x^2+y^2}}=\frac{1}{\sqrt{x^2+y^2}}.$$

$$dz=\frac{x}{y\sqrt{x^2+y^2}+x^2+y^2}dx+\frac{1}{\sqrt{x^2+y^2}}dy.$$

Answer: $dz=\frac{x}{y\sqrt{x^2+y^2}+x^2+y^2}dx+\frac{1}{\sqrt{x^2+y^2}}dy.$

7.91. $z=\ln\cos\frac{x}{y}.$

Solution.

$$dz=z'_xdx+z'_ydy.$$

Let's find the partial derivatives:

$$z'_x=(\ln\cos\frac{x}{y})'_x=\frac{1}{\cos\frac{x}{y}}(\cos\frac{x}{y})'_x=-\frac{1}{\cos\frac{x}{y}}\sin\frac{x}{y}(\frac{x}{y})'_x=$$ $$=-\frac{1}{\cos\frac{x}{y}}\sin\frac{x}{y}\frac{1}{y}.$$

$$z'_y=(\ln\cos\frac{x}{y})'_y=\frac{1}{\cos\frac{x}{y}}(\cos\frac{x}{y})'_y=-\frac{1}{\cos\frac{x}{y}}\sin\frac{x}{y}(\frac{x}{y})'_y=$$ $$=\frac{1}{\cos\frac{x}{y}}\sin\frac{x}{y}\frac{x}{y^2}.$$

$$dz=\left(-\frac{1}{\cos\frac{x}{y}}\sin\frac{x}{y}\frac{1}{y}\right)dx+\left(\frac{1}{\cos\frac{x}{y}}\sin\frac{x}{y}\frac{x}{y^2}\right)dy=-\frac{tg\frac{x}{y}}{y}dx+\frac{xtg\frac{x}{y}}{y^2}.$$

Answer: $dz=-\frac{tg\frac{x}{y}}{y}dx+\frac{xtg\frac{x}y{}}{y^2}dy.$

7.95. Compute approximately $(2.01)^{3.03}$.

Solution.

The desired number will be considered as the value of the function $f(x, y)=x^y$ at $x=x_0+\Delta x, y=y_0+\Delta y,$ where $x_0=2, y_0=3,$ $\Delta x=0.01,$ and $\Delta y=0.03.$ We have:

$$f(2, 3)=2^3=8,$$

$$\Delta f(x, y)\approx d f(x, y)=yx^{y-1}dx+x^y\ln xdy$$

$$\Delta f(2, 3)\approx 3\cdot 2^{3-1}\cdot 0,01+2^3\ln 2\cdot0,03\approx 0,06+0,17=0,23.$$

Therefore, $(2,01)^{3,03}\approx 8+0,23=8,23.$

Answer: $8,23.$

Directional derivative.

If the direction $l$ in the space $Oxyz$ is characterized by direction cosines ${\cos\alpha, \cos\beta, \cos\gamma}$ and the function $u=f(x, y, z)$ is differentiable, then the directional derivative along $l$ is calculated using the formula:

$$\frac{\partial u}{\partial l}=\frac{\partial u}{\partial x}\cos\alpha+\frac{\partial u}{\partial y}\cos\beta+\frac{\partial u}{\partial z}\cos \gamma.$$

The rate of maximum increase of functions at a given point, in magnitude and direction, is determined by the vector - the gradient of the function:

$$grad u=\frac{\partial u}{\partial x}i+\frac{\partial u}{\partial y}j+\frac{\partial u}{\partial z}k,$$the value of which is equal to

$$|grad u|=\sqrt{\left(\frac{\partial u}{\partial x}\right)^2+\left(\frac{\partial u}{\partial y}\right)^2+\left(\frac{\partial u}{\partial z}\right)^2}.$$

Examples: (Demidovich)

3341Find the derivative of the function $z=x^2-y^2$ at the point $M(1, 1)$ in the direction of $l$ that makes an angle $\alpha=\pi/3$ with the positive direction of the $Ox$ axis.

Solution.

The direction $l$ is characterized by direction cosines $\cos\alpha=\cos\pi/3=1/2;$

$\cos\beta=\cos(\pi/2-\pi/3)=\cos\pi/6=\sqrt{3}/2.$

We find the directional derivative using the formula $$\frac{\partial z}{\partial l}=\frac{\partial z}{\partial x}\cos\alpha+\frac{\partial z}{\partial y}\cos\beta.$$

$$z'_x=2x;$$

$$z'_y=-2y;$$

$$\frac{\partial z}{\partial l}=2x\cdot\frac{1}{2}-2y\cdot\frac{\sqrt 3}{2}.$$

$$\frac{\partial z}{\partial l}=2\cdot 1\cdot\frac{1}{2}-2\cdot 1\cdot\frac{\sqrt 3}{2}=1-\sqrt 3.$$

Answer: $1-\sqrt 3.$

3345. Find the derivative of the function $u=xyz$ at the point $M(1, 1, 1)$ in the direction of $l(\cos\alpha, \cos\beta, \cos\gamma)$. What is the magnitude of the gradient of the function at this point?

Solution.

The direction $l$ is characterized by direction cosines.

We find the directional derivative using the formula $$\frac{\partial u}{\partial l}=\frac{\partial u}{\partial x}\cos\alpha+\frac{\partial u}{\partial y}\cos\beta+\frac{\partial u}{\partial z}\cos \gamma.$$

$$u'_x=yz;$$

$$u'_y=xz;$$

$$u'_z=xy,$$ $$\frac{\partial u}{\partial l}=yz\cos\alpha+xz\cos\beta+xy\cos \gamma.$$

$$\frac{\partial u}{\partial l}|_{M(1, 1, 1)}=\cos\alpha+\cos\beta+\cos \gamma.$$

$$grad u=(yz, xz, xy);$$

$$grad u|_{M(1, 1, 1)}=(1, 1, 1);$$

$$|grad u| _{M(1, 1, 1)}=\sqrt{1+1+1}=\sqrt 3$$

Answer: $\frac{\partial u}{\partial l}=\cos\alpha+\cos\beta+\cos\gamma; |grad u|=\sqrt{3}.$

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