Differential

Idea

Differential $df$ is the change in tangent line or tangent plane between 2 given points in the domain.

Where this can be used?

1. Approximate function computation
2. Change of variables in integration and differentiation
3. Explanation for derivative notation (as a quotient of 2 differentials)

Definition

The change in the function $f(x)$ when the independent variable $x$ changes by $\Delta x$ can be written as:

(1)
\begin{align} \Delta f(x) = f(x+\Delta x)-f(x)=A(x) \Delta x + B(x,\Delta x) \end{align}

where $\frac{B}{{\Delta x}}\mathop \to \limits_{\Delta x \to 0} 0$.

Differential
a principal (linear) part $df=A(x)\Delta x$.

Note: The differential is defined for any dependent or independent difference $\Delta x$, but we use it only for infinitely small change $dx$ (differential of variable $x$).

** Example **

(2)
$$f(x)=x^2+3x$$
(3)
\begin{align} f(x+\Delta x)-f(x)=(x+\Delta x)^2+3(x+\Delta x)-x^2-3x=(2x+3)\Delta x + (\Delta x)^2 \end{align}

Thus, $df=(2x+3)\Delta x$.

Alternative definition

Differential Invariance Theorem

(4)
$$df(x)=f'(x)dx$$

where $f'$ is a derivative of$f$, and $dx$ is a differential of dependent or independent variable $x$ (e.g., $x$ can be a function of variable $p$).

Example

(5)
\begin{align} d(x^2+3x)=\frac{d(x^2+3x)}{dx}dx=(2x+3)dx \end{align}

Example
Imagine the cube with all sides of $x=10$. It has a volume $V=10^3=1000$. Suppose each side enlarges by 1 (10%). We want to estimate the change in the volume. Of course, we can do that directly for such a simple problem by:
$\Delta V = (10+1)^3-10^3=331$, but we may try to estimate $V=x^3$ by computing the differential:
$dV=3x^2dx=3*100*1=300$. The approximation is not so good, but it should be better for smaller $dx$.

Differential rules

• Differential of constant:
(9)
$$da=0$$
• Differential of independent variable:
(10)
\begin{align} dx=\Delta x \end{align}
(11)
• Differential of composite function:
(12)
\begin{align} df\left(x(p)\right)=f'(p)dp \end{align}
• Differential of multiplication:
(13)
$$d(uv)=udv+vdu$$
(14)
$$d(uvw)=vwdu+uwdv+uvdw$$
• Differential of division:
(15)
\begin{align} d\left(\frac{u}{v}\right)=\frac{vdu-udv}{v^2} \end{align}

Higher order differentials

Higher order differentials can be computed from lower order by applying the differential repeatedly:

(16)
\begin{align} d^n f=d\left( d^n-1 \right) \end{align}

Theorem
For independent variable $dx$:

(17)
$$d^n f = f^{(n)}dx^n$$

Note: if $x=x(p)$, then for $n>1$ in general $d^n f(x)$ is not invariant and not equal to $d^n f(p)$.

Differential of vector function

If $\vec f(x) = \left( f_1(x),f_2(x),...,f_n(x) \right)$ then

(19)
\begin{align} d \vec f = \vec f'(x)dx = \left( df_1,...,df_n \right) \end{align}

Rules for differential of vector function

• $d\left( g(x) \vec f(x) \right) = g(x) d \vec f(x) + \vec f(x) dg(x)$
• $d\left( \vec g(x) \cdot \vec f(x) \right) = \vec g(x) \cdot d \vec f(x) + d \vec g(x) \cdot \vec f(x)$
• $d\left( \vec g(x) \times \vec f(x) \right) = \vec g(x) \times d \vec f(x) + d \vec g(x) \times \vec f(x)$

where $"\cdot"$ and $"\times"$ are vector operations.

Differential of multivariate function

As in a single variable case, for a function $f(x,y,z)$ the differential is a principal linear part of $\Delta f$:

(20)
\begin{align} \Delta f(x,y,z) = (A\Delta x + B \Delta y + C \Delta z) + \epsilon \end{align}

where $df=A\Delta x + B \Delta y + C \Delta z$ and $\epsilon/ \rho \to 0$ (with $\rho=\sqrt {\Delta {x^2} + \Delta {y^2} + \Delta {z^2}}$) consists of higher order terms.

Theorem

(21)
$$df(x,y,z) = f_x(x,y,z)dx+f_y(x,y,z)dy+f_z(x,y,z)dz$$

where $f_x=\partial f/\partial x, f_y=\partial f/\partial y, f_z=\partial f/\partial z$ ar partial derivatives.

Note: Geometrically, for $f(x,y)$, the value of $df$ is the difference measured as if the function $f$ was replaced by its tangent plane (see image at the top of this page).

Higher order differentials of multivariate function

For $f(x,y,z)$:

(22)
$$d^n = d(d^{n-1}f)$$

For example for $f(x,y)$ with independent variables $x,y$:

(23)
\begin{align} d^n f = \left( dx \frac{\partial}{\partial x} + dy \frac{\partial}{\partial y} \right) ^n f \end{align}

Inverse problem of differential existence

We want to know for which P and Q the expression $P(x,y)dx+Q(x,y)dy$ is a differential of some function $f(x,y)$ .

Theorem

1. $P(x,y)dx+Q(x,y)dy$ is a differential of some function $f(x,y)$ if and only if $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$ (or $\nabla \times (P,Q) = 0$).
2. $P(x,y,z)dx+Q(x,y,z)dy+R(x,y,z)dz$ can be a differential of some function $f(x,y,z)$ if $\nabla \times (P,Q,R) = 0$. The same can be extended for more variables.

Differential in polar coordinates

(24)
\begin{align} dx = d(r cos(\phi)) = cos(\phi) dr - r sin(\phi) d \phi \end{align}
(25)
\begin{align} dy = d(r sin(\phi)) = sin(\phi) dr + r \cos (\phi) d \phi \end{align}

Annotated references

The best and simplest explanations without proofs are in [1]. Some easy proofs can be found in [2].

Bibliography
1. М.Я. Выгодский, "Справочник по высшей математике," НАУКА, 1977.
2. Э. ФРИД, И. ПАСТОР, И. РЕЙМАН, П. РЕВЕС, И. РУЖА, "МАЛАЯ МАТЕМАТИЧЕСКАЯ ЭНЦИКЛОПЕДИЯ," AKADEMIAI KIADO, 1976.
page revision: 17, last edited: 22 Aug 2011 05:40