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. $$\underset{\delta x_k\to 0}{lim}\frac{f(x_1,...,x_k+\delta x_k,...,x_n)-f(x_1,...,x_k,...,x_n)}{\delta x_k}.$$

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 $f_{x_k}^{\prime }(x_1,...,x_n)$.

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: $$\frac{\partial }{\partial x_k}\left(\frac{\partial u}{\partial x_k}\right)=\frac{{\partial }^{2}u}{\partial x_k^2}=f_{x_k{x}_k}^{″}(x_1,x_2,...,x_k,...,x_n).$$ $$\frac{\partial }{\partial x_l}\left(\frac{\partial u}{\partial x_k}\right)=\frac{{\partial }^{2}u}{\partial x_k\partial x_l}=f_{x_k{x}_l}^{″}(x_1,x_2,...,x_k,...,x_l,...,x_n).$$ 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. $z=x^5+y^5-5x^3{y}^3.$

Solution.

$$z_x^{\prime }=(x^5+y^5-5x^3{y}^3{)}_x^{\prime }=5x^4-15x^2{y}^3;$$

$$z_y^{\prime }=(x^5+y^5-5x^3{y}^3{)}_y^{\prime }=5y^4-15x^3{y}^2;$$

$$z_{xx}^{\prime }=(5x^4-15x^2{y}^3{)}_x^{\prime }=20x^3-30x{y}^3;$$

$$z_{xy}^{\prime }=(5x^4-15x^2{y}^3{)}_y^{\prime }=-45x^2{y}^2;$$

$$z_{yy}^{\prime }=(5y^4-15x^3{y}^2{)}_y^{\prime }=20y^3-30x^{3}y;$$

$$z_{yx}^{\prime }=(5y^4-15x^3{y}^2{)}_x^{\prime }=-45x^2{y}^2.$$

Answer: $z_x^{\prime }=5x^4-15x^2{y}^3;$ $z_y^{\prime }=5y^4-15x^3{y}^2;$ $z_{xx}^{\prime }=20x^3-30x{y}^3;$ $z_{xy}^{\prime }=-45x^2{y}^2;$ $z_{yy}^{\prime }=20y^3-30x^{3}y;$ $z_{yx}^{\prime }=-45x^2{y}^2.$

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

Solution.

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

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

$$z_{xx}^{\prime }={\left(\frac{y^3}{(x^2+y^2{)}^{3/2}}\right)}_x^{\prime }=-\frac{3}{2}{y}^{3}2x(x^2+y^2{)}^{-5/2}=-3y^{3}x(x^2+y^2{)}^{-5/2};$$

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

$$z_{yy}^{\prime }={\left(\frac{x^3}{(x^2+y^2{)}^{3/2}}\right)}_y^{\prime }=-\frac{3}{2}{x}^{3}2y(x^2+y^2{)}^{-5/2}=-3x^{3}y(x^2+y^2{)}^{-5/2};$$

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

Answer: $z_x^{\prime }=\frac{y^3}{(x^2+y^2)\sqrt{x^2+y^2}};$ $z_y^{\prime }=\frac{x^3}{(x^2+y^2)\sqrt{x^2+y^2}};$ $z_{xx}^{\prime }=-3y^{3}x(x^2+y^2{)}^{-5/2};$ $z_{xy}^{\prime }=\frac{3y^2{x}^2}{(x^2+y^2{)}^{5/2}};$ $z_{yy}^{\prime }=-3x^{3}y(x^2+y^2{)}^{-5/2};$ $z_{yx}^{\prime }=\frac{3x^2{y}^2}{(x^2+y^2{)}^{5/2}}.$

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

Solution.

$$z_x^{\prime }={(\ln (x^2+y^2))}_x^{\prime }=\frac{2x}{x^2+y^2};$$

$$z_y^{\prime }={(\ln (x^2+y^2))}_y^{\prime }=\frac{2y}{x^2+y^2};$$

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

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

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

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

Answer: $z_x^{\prime }=\frac{2x}{x^2+y^2};$ $z_y^{\prime }=\frac{2y}{x^2+y^2};$ $z_{xx}^{\prime }=\frac{2(-x^2+y^2)}{x^2+y^2};$ $z_{xy}^{\prime }=\frac{-4xy}{(x^2+y^2{)}^2};$ $z_{yy}^{\prime }=\frac{2(x^2-y^2)}{x^2+y^2};$ $z_{yx}^{\prime }=\frac{-4xy}{(x^2+y^2{)}^2}.$

4. Find $f_x^{\prime }(3,2),f_y^{\prime }(3,2),$ $f_{xx}^{\prime }(3,2),f_{xy}^{\prime }(3,2),$ $f_{yy}^{\prime }(3,2),$ if $f(x,y)=x^{3}y+x{y}^2-2x+3y-1.$

Solution.

Let's find the partial derivatives:

$$f_x^{\prime }={(x^{3}y+x{y}^2-2x+3y-1)}_x^{\prime }=3x^{2}y+y^2;$$

$$f_y^{\prime }={(x^{3}y+x{y}^2-2x+3y-1)}_y^{\prime }=x^3+2xy+3;$$

$$f_{xx}^{\prime }={(3x^{2}y+y^2)}_x^{\prime }=6xy;$$

$$f_{xy}^{\prime }={(3x^{2}y+y^2)}_y^{\prime }=3x^2+2y;$$

$$f_{yy}^{\prime }={(x^3+2xy+3)}_x^{\prime }=2x.$$

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

$$f_x^{\prime }(3,2)=(3x^{2}y+y^2){|}_{(3,2)}=54+4=58;$$

$$f_y^{\prime }(3,2)=(x^3+2xy+3){|}_{(3,2)}=27+12+3=42;$$

$$f_{xx}^{\prime }(3,2)=6xy{|}_{(3,2)}=36;$$

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

$$f_{yy}^{\prime }(3,2)=2x{|}_{(3,2)}=6.$$

Answer: $f_x^{\prime }(3,2)=58,f_y^{\prime }(3,2)=42,$ $f_{xx}^{\prime }(3,2)=36,f_{xy}^{\prime }(3,2)=31,$ $f_{yy}^{\prime }(3,2)=4.$

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:

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

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

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

Answer: Proven.

For the differential of the function $u=f(x_1,x_2,...,x_n)$, the formula applies: $$du=\frac{\partial u}{\partial x_1}d{x}_1+\frac{\partial u}{\partial x_2}d{x}_2+...+\frac{\partial u}{\partial x_n}d{x}_n.$$

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 $\rho =\sqrt{\mathrm{\Delta }{x}_1^2+\mathrm{\Delta }{x}_2^2+...+\mathrm{\Delta }{x}_n^2}$, for a differentiable function $u=f(x_1,x_2,...,x_n)$, approximate equalities occur: $$\mathrm{\Delta }u\approx du,$$ $$f(x_1+\mathrm{\Delta }{x}_1,x_2+\mathrm{\Delta }{x}_2,...,x_n+\mathrm{\Delta }{x}_n)\approx f(x_1,x_2,...,x_n)+df(x_1,x_2,...,x_n).$$

Examples:

Find differentials of functions:

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

Solution.

$$dz=z_x^{\prime }dx+z_y^{\prime }dy.$$

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

$$z_y^{\prime }=(\ln (y+\sqrt{x^2+y^2}){)}_y^{\prime }=\frac{1}{y+\sqrt{x^2+y^2}}(y+\sqrt{x^2+y^2}{)}_y^{\prime }=$$ $$=\frac{1}{y+\sqrt{x^2+y^2}}(1+\frac{1}{2\sqrt{x^2+y^2}}(x^2+y^2{)}_y^{\prime })=$$ $$=\frac{1}{y+\sqrt{x^2+y^2}}(1+\frac{1}{2\sqrt{x^2+y^2}}2y)=\frac{1}{y+\sqrt{x^2+y^2}}+\frac{y}{\sqrt{x^2+y^2}(y+\sqrt{x^2+y^2})}=$$ $$=\frac{\sqrt{x^2+y^2}+y}{(y+\sqrt{x^2+y^2})\sqrt{x^2+y^2}}=\frac{1}{\sqrt{x^2+y^2}}.$$

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

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.

$$dz=z_x^{\prime }dx+z_y^{\prime }dy.$$

Let's find the partial derivatives:

$$z_x^{\prime }=(\ln \cos \frac{x}{y}{)}_x^{\prime }=\frac{1}{\cos \frac{x}{y}}(\cos \frac{x}{y}{)}_x^{\prime }=-\frac{1}{\cos \frac{x}{y}}\sin \frac{x}{y}(\frac{x}{y}{)}_x^{\prime }=$$ $$=-\frac{1}{\cos \frac{x}{y}}\sin \frac{x}{y}\frac{1}{y}.$$

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

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

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 $x=x_0+\mathrm{\Delta }x,y=y_0+\mathrm{\Delta }y,$ where $x_0=2,y_0=3,$ $\mathrm{\Delta }x=0.01,$ and $\mathrm{\Delta }y=0.03.$ We have:

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

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

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

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:

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

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

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

$$z_x^{\prime }=2x;$$

$$z_y^{\prime }=-2y;$$

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

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

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

$$u_x^{\prime }=yz;$$

$$u_y^{\prime }=xz;$$

$$u_z^{\prime }=xy,$$ $$\frac{\partial u}{\partial l}=yz\cos \alpha +xz\cos \beta +xy\cos \gamma .$$

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

$$gradu=(yz,xz,xy);$$

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

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

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

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