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: $$\frac{\partial}{\partial x_k}\left(\frac{\partial u}{\partial x_k}\right)=\frac{\partial^2u}{\partial x^2_k}=f''_{x_kx_k}(x_1, x_2, ..., x_k, ..., x_n).$$ $$\frac{\partial}{\partial x_l}\left(\frac{\partial u}{\partial x_k}\right)=\frac{\partial^2u}{\partial x_k\partial x_l}=f''_{x_kx_l}(x_1, x_2, ..., x_k, ..., x_l, ..., x_n).$$ and so on.
Similarly, partial derivatives of order higher than the second are defined and denoted.
The second-order differential $d^2f$ 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 $dx_1, dx_2, ..., dx_n:$$$d^2u=d(du).$$
Similarly, the third-order differential is defined as:$$d^3u=d(d^2u).$$
In general, $$d^mu=d(d^{m-1}u).$$
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 $$d^mu=\left(\frac{\partial}{\partial x_1}dx_1+\frac{\partial}{\partial x_2}dx_2+...+\frac{\partial}{\partial x_n}dx_n\right)^mu,$$ 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 $$d^2z=\frac{\partial^2 z}{\partial x^2}dx^2+2\frac{\partial^2 z}{\partial x\partial y}dxdy+\frac{\partial^2 z}{\partial y^2}dy^2,$$
$$d^3z=\frac{\partial^3 z}{\partial x^3}dx^3+3\frac{\partial^3 z}{\partial x^2\partial y}dx^2dy++3\frac{\partial^3 z}{\partial x\partial y^2}dxdy^2+\frac{\partial^3 z}{\partial y^3}dy^3.$$
Examples.
Find the first and second-order differentials of the following functions: ($x, y, z$ are independent variables):
1. $z=x^3+3x^2y-y^3.$
Solution.
$$dz=z'_xdx+z'_ydy.$$
$$z'_x=(x^3+3x^2y-y^3)'_x=3x^2+6xy;$$
$$z'_y=(x^3+3x^2y-y^3)'_y=3x^2-3y^2.$$
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).: $$d^2z=d((3x^2+6xy)dx+(3x^2-3y^2)dy)=d(3x^2+6xy)dx+d(3x^2-3y^2)dy=$$ $$=(6xdx+6xdy+6ydx)dx+(6xdx-6ydy)dy=6((x+y)dx^2+2xdxdy-ydy^2).$$Answer: $dz=(3x^2+6xy)dx+(3x^2-3y^2)dy;$ $d^2 z=6((x+y)dx^2+2xdxdy-ydy^2).$
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):$$d^2u=d((y+z)dx+(x+z)dy+(x+y)dz)=$$ $$=(dy+dz)dx+(dx+dz)dy+(dx+dy)dz=2(dxdy+dxdz+dydz).$$
Answer: $du=(y+z)dx+(x+z)dy+(x+y)dz;$ $d^2 u=2(dxdy+dxdz+dydz).$
3. Find $d^3z$ if $z = e^y \sin x.$
Solution.
$$dz=d(e^y\sin x)=e^y\cos xdx+\sin xe^y dy.$$ Differentiate again, taking into account that $dx$ and $dy$ do not depend on $x$ and $y$ (i.e., treating $dx$ and $dy$ as constants).: $$d^2z=d(e^y\cos xdx+\sin xe^y dy)=$$ $$=-\sin x e^ydxdx+\cos x e^y dydx+e^y\cos xdxdy+\sin xe^ydydy=$$ $$=-\sin x e^y dx^2+2e^y\cos xdxdy+\sin xe^y dy^2.$$
$$d^3z=d(-\sin x e^y dx^2+2e^y\cos xdxdy+\sin xe^y dy^2)=$$ $$=(-\cos xe^ydx^2-2e^y\sin xdxdy+\cos xe^ydy^2)dx+$$ $$+(-\sin xe^ydx^2+2e^y\cos xdxdy+\sin xe^ydy^2)dy=$$ $$=-\cos xe^ydx^3+-3e^y\sin xdx^2dy+3\cos xy^ydxdy^2+\sin xe^ydy^3.$$
Answer: $d^3z = -\cos xe^y dx^3 - 3e^y \sin x dx^2dy + 3\cos xy^y dx dy^2 + \sin xe^y dy^3.$
4. Find $d^6u,$ if $u = \ln(x+y+z).$
Solution.
We will use the formula
$$d^mu=\left(\frac{\partial}{\partial x_1}dx_1+\frac{\partial}{\partial x_2}dx_2+...+\frac{\partial}{\partial x_n}dx_n\right)^mu,$$
$$\frac{\partial u}{\partial x}=\frac{\partial u}{\partial y}=\frac{\partial u}{\partial z}=\frac{1}{x+y+z};$$
$$\frac{\partial^2 u}{\partial x^2}=\frac{\partial^2 u}{\partial y^2}=\frac{\partial^2 u}{\partial z^2}=\frac{\partial^2u}{\partial x\partial y}=\frac{\partial^2u}{\partial y\partial x}=\frac{\partial^2u}{\partial x\partial z}=\frac{\partial^2u}{\partial z\partial x}=\frac{\partial^2u}{\partial y\partial z}=\frac{\partial^2u}{\partial z\partial y}=$$ $$=-\frac{1}{(x+y+z)^2};$$
...
$$\frac{\partial^6 u}{\partial x^6}=\frac{\partial^6 u}{\partial y^6}=\frac{\partial^6 u}{\partial z^6}=\frac{\partial^6u}{\partial x^5\partial y}=\frac{\partial^6u}{\partial x^4\partial y^2}=...\frac{\partial^6u}{\partial x\partial y^5}=\frac{\partial^6u}{\partial x^5\partial z}=...=$$ $$=\frac{\partial^6u}{\partial x\partial z^5}=\frac{\partial^6u}{\partial y^5\partial z}=...=\frac{\partial^6 u}{\partial y\partial z^5}=(-1)^55!\frac{1}{(x+y+z)^6}=\frac{5!}{(x+y+z)^6}.$$
From here we have $$d^6(\ln(x+y+z))=\left(\frac{\partial}{\partial x}dx+\frac{\partial}{\partial y}dy+\frac{\partial}{\partial z}dz\right)^6(\ln(x+y+z))=$$ $$=-\frac{5!}{(x+y+z)^6}(dx+dy+dz)^6.$$
Answer: $d^6u=-\frac{5!}{(x+y+z)^6}(dx+dy+dz)^6.$
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 $d^3u,$ if $u=x^3+y^3+z^3-3xyz..$
6. Find $d^mu,$ if $u=e^{ax+by+cz}.$
Tags: Higher-order derivatives, calculus, derivative table, detivative, differentials, mathematical analysis
