Partial derivatives, differential, directional derivative, gradient.

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:

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

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

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

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

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

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

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

5. 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:

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

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

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

8. 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:

1. Find 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: $1-\sqrt 3.$

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

Tags: Partial derivatives, calculus, derivative, differential, directional derivative, limDirectional derivative, mathematical analysis