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 D_x f(x).

Linear Combination
f(x) = a·g(x) + b·h(x)
df(x) = d(a·g(x) + b·h(x)) =
= D_x(a·g(x) + b·h(x))·dx =
= (a·D_xg(x) + b·D_xh(x))·dx =
= a·D_xg(x)·dx + b·D_xh(x)·dx =
= a·dg(x) + b·dh(x)

Product
f(x) = g(x)·h(x)
df(x) = d(g(x)·h(x)) =
= D_x(g(x)·h(x))·dx =
= (D_xg(x)·h(x) +
+ g(x)·D_xh(x))·dx =
= D_xg(x)·dx·h(x) +
+ g(x)·D_xh(x)·dx =
= dg(x)·h(x)+g(x)·dh(x)

Reciprocal
f(x) = 1/g(x)
df(x) = d(1/g(x)) =
= D_x(1/g(x))·dx =
= −[1/g²(x)]·D_xg(x)·dx =
= −[1/g²(x)]·dg(x)

Compound
f(x) = g(h(x))
df(x) = d(g(h(x))) =
= D_xg(h(x))·dx =
= D_yg(y)·D_xh(x)·dx =
= D_yg(y)·dh(x)
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) = D_x f(x)·dx =
= D_xsin(ln(x))·dx =
using the chain rule of differentiation of a compound function
= cos(ln(x))·(1/x)·dx
On the other hand, let's see what our expression for differential of a compound function gives.
df(x) = d(g(h(x))) =
= D_yg(y)·dh(x) =
= D_ysin(y)·dln(x) =
substituting y with ln(x).
= cos(y)·(1/x)·dx =
= cos(ln(x))·(1/x)·dx
As you see, the result is the same as with direct computing the differential.

Implicit
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) = D_yg(y)·dh(x),
we conclude that
dh(x) = df(x)/D_yg(y)
substituting y with h(x).
For example, we need to find dln(x) without the knowledge that D_xln(x) = 1/x, which would immediately give us
dln(x) = (1/x)·dx
Consider an equality
x = e^ln(x)
If functions are equal, their differentials at any point are also equal.
dx = de^ln(x)
Now we can use the expression for a differential of a compound function e^ln(x) where g(x)=e^xand h(x)=ln(x).
dx = de^ln(x) = D_ye^y·dln(x)
substituting y with ln(x).
Hence,
dx = e^y·dln(x) =
= e^ln(x)·dln(x) = x·dln(x)
Finally,
dln(x) = (1/x)·dx
which corresponds to the result of direct calculations if we use the known expression D_xln(x)=1/x.