Higher-order derivatives and differentials.

The second-order partial derivatives of the function $u=f(x_1,x_2,...,x_n)$ are the partial derivatives of its first-order partial derivatives. The second-order derivatives are denoted as follows: $$\frac{\partial }{\partial x_k}\left(\frac{\partial u}{\partial x_k}\right)=\frac{{\partial }^{2}u}{\partial x_k^2}=f_{x_k{x}_k}^{″}(x_1,x_2,...,x_k,...,x_n).$$ $$\frac{\partial }{\partial x_l}\left(\frac{\partial u}{\partial x_k}\right)=\frac{{\partial }^{2}u}{\partial x_k\partial x_l}=f_{x_k{x}_l}^{″}(x_1,x_2,...,x_k,...,x_l,...,x_n).$$ and so on.

Similarly, partial derivatives of order higher than the second are defined and denoted.

The second-order differential $d^{2}f$ of the function $u=f(x_1,x_2,...,x_n)$ is called the differential of its first-order differential, considered as a function of the variables $x_1,x_2,...,x_n$ with fixed values of $d{x}_1,d{x}_2,...,d{x}_n:$$$d^{2}u=d(du).$$

Similarly, the third-order differential is defined as:$$d^{3}u=d(d^{2}u).$$

In general, $$d^{m}u=d(d^{m-1}u).$$

The $m$-th order differential of the function $u=f(x_1,x_2,...,x_n),$ where $x_1,x_2,..,x_n-$ are independent variables, is expressed by the symbolic formula $$d^{m}u={(\frac{\partial }{\partial x_1}d{x}_1+\frac{\partial }{\partial x_2}d{x}_2+...+\frac{\partial }{\partial x_n}d{x}_n)}^{m}u,$$ which is formally expanded according to the binomial theorem.

For example, in the case of the function $z=f(x,y)$ of two independent variables, the formulas for the second and third-order differentials are valid $$d^{2}z=\frac{{\partial }^{2}z}{\partial x^2}d{x}^2+2\frac{{\partial }^{2}z}{\partial x\partial y}dxdy+\frac{{\partial }^{2}z}{\partial y^2}d{y}^2,$$

$$d^{3}z=\frac{{\partial }^{3}z}{\partial x^3}d{x}^3+3\frac{{\partial }^{3}z}{\partial x^2\partial y}d{x}^{2}dy++3\frac{{\partial }^{3}z}{\partial x\partial y^2}dxd{y}^2+\frac{{\partial }^{3}z}{\partial y^3}d{y}^3.$$

Examples.

Find the first and second-order differentials of the following functions: ($x,y,z$ are independent variables):

1. $z=x^3+3x^{2}y-y^3.$

Solution.

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

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

$$z_y^{\prime }=(x^3+3x^{2}y-y^3{)}_y^{\prime }=3x^2-3y^2.$$

Thus

$$dz=(3x^2+6xy)dx+(3x^2-3y^2)dy.$$

Differentiate a second time, considering that $dx$ and $dy$ are independent of $x$ and $y$ (i.e., treating $dx$ and $dy$ as constants).: $$d^{2}z=d((3x^2+6xy)dx+(3x^2-3y^2)dy)=d(3x^2+6xy)dx+d(3x^2-3y^2)dy=$$ $$=(6xdx+6xdy+6ydx)dx+(6xdx-6ydy)dy=6((x+y)d{x}^2+2xdxdy-yd{y}^2).$$Answer: $dz=(3x^2+6xy)dx+(3x^2-3y^2)dy;$ $d^{2}z=6((x+y)d{x}^2+2xdxdy-yd{y}^2).$

2.$u=xy+yz+zx.$

Solution.

$$du=d(xy+yz+zx)=xdy+ydx+ydz+zdy+zdx+xdz=$$ $$=(y+z)dx+(x+z)dy+(x+y)dz.$$

Differentiate a second time, taking into account that $dx,,,dy$, and $dz$ do not depend on $x,,,y$, and $z$ (i.e., treating $dx,,dy$, and $dz$ as constants):$$d^{2}u=d((y+z)dx+(x+z)dy+(x+y)dz)=$$ $$=(dy+dz)dx+(dx+dz)dy+(dx+dy)dz=2(dxdy+dxdz+dydz).$$

Answer: $du=(y+z)dx+(x+z)dy+(x+y)dz;$ $d^{2}u=2(dxdy+dxdz+dydz).$

3. Find $d^{3}z$ if $z=e^y\sin x.$

Solution.

$$dz=d(e^y\sin x)=e^y\cos xdx+\sin x{e}^{y}dy.$$ Differentiate again, taking into account that $dx$ and $dy$ do not depend on $x$ and $y$ (i.e., treating $dx$ and $dy$ as constants).: $$d^{2}z=d(e^y\cos xdx+\sin x{e}^{y}dy)=$$ $$=-\sin x{e}^{y}dxdx+\cos x{e}^{y}dydx+e^y\cos xdxdy+\sin x{e}^{y}dydy=$$ $$=-\sin x{e}^{y}d{x}^2+2e^y\cos xdxdy+\sin x{e}^{y}d{y}^2.$$

$$d^{3}z=d(-\sin x{e}^{y}d{x}^2+2e^y\cos xdxdy+\sin x{e}^{y}d{y}^2)=$$ $$=(-\cos x{e}^{y}d{x}^2-2e^y\sin xdxdy+\cos x{e}^{y}d{y}^2)dx+$$ $$+(-\sin x{e}^{y}d{x}^2+2e^y\cos xdxdy+\sin x{e}^{y}d{y}^2)dy=$$ $$=-\cos x{e}^{y}d{x}^3+-3e^y\sin xd{x}^{2}dy+3\cos x{y}^{y}dxd{y}^2+\sin x{e}^{y}d{y}^3.$$

Answer: $d^{3}z=-\cos x{e}^{y}d{x}^3-3e^y\sin xd{x}^{2}dy+3\cos x{y}^{y}dxd{y}^2+\sin x{e}^{y}d{y}^3.$

4. Find $d^{6}u,$ if $u=\ln (x+y+z).$

Solution.

We will use the formula

$$d^{m}u={(\frac{\partial }{\partial x_1}d{x}_1+\frac{\partial }{\partial x_2}d{x}_2+...+\frac{\partial }{\partial x_n}d{x}_n)}^{m}u,$$

$$\frac{\partial u}{\partial x}=\frac{\partial u}{\partial y}=\frac{\partial u}{\partial z}=\frac{1}{x+y+z};$$

$$\frac{{\partial }^{2}u}{\partial x^2}=\frac{{\partial }^{2}u}{\partial y^2}=\frac{{\partial }^{2}u}{\partial z^2}=\frac{{\partial }^{2}u}{\partial x\partial y}=\frac{{\partial }^{2}u}{\partial y\partial x}=\frac{{\partial }^{2}u}{\partial x\partial z}=\frac{{\partial }^{2}u}{\partial z\partial x}=\frac{{\partial }^{2}u}{\partial y\partial z}=\frac{{\partial }^{2}u}{\partial z\partial y}=$$ $$=-\frac{1}{(x+y+z{)}^2};$$

...

$$\frac{{\partial }^{6}u}{\partial x^6}=\frac{{\partial }^{6}u}{\partial y^6}=\frac{{\partial }^{6}u}{\partial z^6}=\frac{{\partial }^{6}u}{\partial x^5\partial y}=\frac{{\partial }^{6}u}{\partial x^4\partial y^2}=...\frac{{\partial }^{6}u}{\partial x\partial y^5}=\frac{{\partial }^{6}u}{\partial x^5\partial z}=...=$$ $$=\frac{{\partial }^{6}u}{\partial x\partial z^5}=\frac{{\partial }^{6}u}{\partial y^5\partial z}=...=\frac{{\partial }^{6}u}{\partial y\partial z^5}=(-1{)}^{5}5!\frac{1}{(x+y+z{)}^6}=\frac{5!}{(x+y+z{)}^6}.$$

From here we have $$d^6(\ln (x+y+z))={(\frac{\partial }{\partial x}dx+\frac{\partial }{\partial y}dy+\frac{\partial }{\partial z}dz)}^6(\ln (x+y+z))=$$ $$=-\frac{5!}{(x+y+z{)}^6}(dx+dy+dz{)}^6.$$

Answer: $d^{6}u=-\frac{5!}{(x+y+z{)}^6}(dx+dy+dz{)}^6.$

Homework.

Find the first and second-order differentials of the following functions: ($x,y,z$ are independent variables):

1. $z=\frac{y}{x}-\frac{x}{y}.$

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

3. $z=(x+y)e^{xy}.$

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

5 Find $d^{3}u,$ if $u=x^3+y^3+z^3-3xyz..$

6. Find $d^{m}u,$ if $u=e^{ax+by+cz}.$

Tags: Higher-order derivatives, calculus, derivative table, detivative, differentials, mathematical analysis