*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)***d*=

**f(x)***d*

**(a·g(x) + b·h(x)) =**

=D=

_{x}**(a·g(x) + b·h(x))·**d**x =**

= (a·D= (a·

_{x}**g(x) + b·**D_{x}**h(x))·**d**x =**

= a·D= a·

_{x}**g(x)·**d**x + b·**D_{x}**h(x)·**d**x =**

= a·d= a·

**g(x) + b·**d**h(x)***Product*

**f(x) = g(x)·h(x)***d*=

**f(x)***d*

**(g(x)·h(x)) =**

=D=

_{x}**(g(x)·h(x))·**d**x =**

= (D= (

_{x}**g(x)·h(x) +**

+ g(x)·D+ g(x)·

_{x}**h(x))·**d**x =**

=D=

_{x}**g(x)·**d**x·h(x) +**

+ g(x)·D+ g(x)·

_{x}**h(x)·**d**x =**

=d=

**g(x)·h(x)+g(x)·**d**h(x)***Reciprocal*

**f(x) = 1/g(x)***d*=

**f(x)***d*

**(1/g(x)) =**

=D=

_{x}**(1/g(x))·**d**x =**

= −= −

**[**

*1/g²(x)*]*·**D*

_{x}**g(x)·**d**x =**

= −= −

**[**

*1/g²(x)*]*·**d*

**g(x)***Compound*

**f(x) = g(h(x))***d*

**f(x)**= d**(g(h(x))) =**

=D=

_{x}**g(h(x))·**d**x =**

=D=

_{y}**g(y)·**D_{x}**h(x)·**d**x =**

=D=

_{y}**g(y)·**d**h(x)**substituting

*with*

**y***.*

**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

*and multiplying it by*

**f(x)***d*.

**x***d*

**f(x) =**D_{x}**f(x)·**d**x =**=

*D*

_{x}**sin(ln(x))·**d**x =**using the chain rule of differentiation of a compound function

=

**cos(ln(x))·(1/x)·**d**x**On the other hand, let's see what our expression for differential of a compound function gives.

*d*=

**f(x)**= d**(g(h(x))) =**

=D=

_{y}**g(y)·**d**h(x)**=

*D*=

_{y}**sin(y)·**d**ln(x)**substituting

*with*

**y***.*

**ln(x)**=

*=*

**cos(y)·(1/x)·**d**x**=

**cos(ln(x))·(1/x)·**d**x**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,

*and we need to find*

**f(x) = g(h(x))***d*.

**h(x)**Since

*d*,

**f(x)**= D_{y}**g(y)·**d**h(x)**we conclude that

*d*

**h(x) =**d**f(x)/**D_{y}**g(y)**substituting

*with*

**y***.*

**h(x)**For example, we need to find

*d*without the knowledge that

**ln(x)***D*

_{x}**ln(x) = 1/x***d*

**ln(x) = (1/x)·**d**x**Consider an equality

**x = e**^{ln(x)}If functions are equal, their differentials at any point are also equal.

*d*

**x =**d**e**^{ln(x)}Now we can use the expression for a differential of a compound function

*where*

**e**^{ln(x)}*and*

**g(x)=e**^{x}*.*

**h(x)=ln(x)***d*

**x**= d**e**D^{ln(x)}=_{y}**e**d^{y}·**ln(x)**substituting

*with*

**y***.*

**ln(x)**Hence,

*d*=

**x**= e^{y}·d**ln(x)**=

**e**d^{ln(x)}·**ln(x) = x·**d**ln(x)**Finally,

*d*

**ln(x) = (1/x)·**d**x**which corresponds to the result of direct calculations if we use the known expression

*D*.

_{x}**ln(x)=1/x**
## No comments:

Post a Comment