Tuesday, January 17, 2017

Unizor - Derivatives - Properties of Differential of a Function

Notes to a video lecture on http://www.unizor.com

Properties of Differential

Properties of differential of a function resemble properties of derivative and immediately follow from them.
Below are the proofs of all these properties, using Euler's notation for derivative Dx f(x).

Linear Combination
f(x) = a·g(x) + b·h(x)
df(x) = d(a·g(x) + b·h(x)) =
Dx(a·g(x) + b·h(x))·dx =
= (a·
Dxg(x) + b·Dxh(x))·dx =
= a·
Dxg(x)·dx + b·Dxh(x)·dx =
= a·
dg(x) + b·dh(x)

f(x) = g(x)·h(x)
df(x) = d(g(x)·h(x)) =
Dx(g(x)·h(x))·dx =
= (
Dxg(x)·h(x) +
+ g(x)·
Dxh(x))·dx =
Dxg(x)·dx·h(x) +
+ g(x)·
Dxh(x)·dx =

f(x) = 1/g(x)
df(x) = d(1/g(x)) =
Dx(1/g(x))·dx =
= −
[1/g²(x)]·Dxg(x)·dx =
= −

f(x) = g(h(x))
df(x) = d(g(h(x))) =
Dxg(h(x))·dx =
Dyg(y)·Dxh(x)·dx =

substituting y with h(x).
Let's illustrate this rule for compound functions.
g(x) = sin(x)
h(x) = ln(x)
f(x) = g(h(x)) = sin(ln(x))
First, let's calculate the differential directly by taking a derivative from f(x) and multiplying it by dx.
df(x) = Dx f(x)·dx =
Dxsin(ln(x))·dx =
using the chain rule of differentiation of a compound function
On the other hand, let's see what our expression for differential of a compound function gives.
df(x) = d(g(h(x))) =
Dysin(y)·dln(x) =
substituting y with ln(x).
cos(y)·(1/x)·dx =
As you see, the result is the same as with direct computing the differential.

Implicitly defined differential can be calculated using the above rule for differential of a compound function.
Assume, f(x) = g(h(x)) and we need to find dh(x).
Since df(x) = Dyg(y)·dh(x),
we conclude that
dh(x) = df(x)/Dyg(y)
substituting y with h(x).
For example, we need to find dln(x) without the knowledge that Dxln(x) = 1/x, which would immediately give us
dln(x) = (1/x)·dx
Consider an equality
x = eln(x)
If functions are equal, their differentials at any point are also equal.
dx = deln(x)
Now we can use the expression for a differential of a compound function eln(x) where g(x)=exand h(x)=ln(x).
dx = deln(x) = Dyey·dln(x)
substituting y with ln(x).
dx = ey·dln(x) =
eln(x)·dln(x) = x·dln(x)
dln(x) = (1/x)·dx
which corresponds to the result of direct calculations if we use the known expression Dxln(x)=1/x.

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