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

__Derivative Properties -__

Product of Two Functions

Product of Two Functions

Our purpose is to express the derivative of a product of two functions in terms of derivatives of each of them.

Assume that two real functions

**and**

*f(x)***are defined and**

*g(x)**differentiable*(that is, their derivatives exist) on certain interval with derivatives, correspondingly,

*f**and*

^{ I}**(x)**

*g**.*

^{I}**(x)**Let's determine the derivative of their product

*h(x) = f(x)·g(x)*The increment of function

**is**

*h(x)*Δ

**=**

*h(x)***Δ**

*h(x+***Δ**

*x)−h(x) =*

= [f(x+= [f(x+

**Δ**

*x)·g(x+***Δ**

*x)] −*

− [f(x)·g(x)] =

= [f(x+− [f(x)·g(x)] =

= [f(x+

**Δ**

*x)·g(x+***Δ**

*x)] −*

− [f(x+− [f(x

**Δ**

*x)·g(x)] +*

+ [f(x++ [f(x

**Δ**

*x)·g(x)] −*

− [f(x)·g(x)] =

= f(x+− [f(x)·g(x)] =

= f(x+

**Δ**

*x)·[g(x+***Δ**

*x)−g(x)] +*

+ g(x)·[f(x++ g(x)·[f(x

**Δ**

*x)−f(x)] =*

= f(x+= f(x+

**Δ**

*x)·***Δ**

*g(x) + g(x)·*

*f(x)*Next operations to find a derivative are: dividing the function increment Δ

**by an increment of an argument Δ**

*h(x)***and going to a limit as Δ**

*x***.**

*x→0*Δ

**Δ**

*h(x)/***Δ**

*x =*

= f(x+= f(x+

**Δ**

*x)·***Δ**

*g(x)/***Δ**

*x +*

+ g(x)·+ g(x)·

**Δ**

*f(x)/*

*x*Since both functions

**and**

*f(x)***are differentiable, there is a limit of Δ**

*g(x)***Δ**

*f(x)/***and Δ**

*x***Δ**

*g(x)/***as Δ**

*x***. These limits are, correspondingly,**

*x→0*

*f**and*

^{ I}**(x)**

*g**.*

^{I}**(x)**At the same time

**Δ**

*f(x+*

*x) → f(x)*as Δ

**.**

*x→0*Recall the properties of limits:

if two sequences have limits

*L*and

*M*, then their sum has a limit

*L+M*and their product has a limit

*L·M*.

Therefore,

Δ

**Δ**

*h(x)/***→**

*x*

*f(x)·g*

^{I}**(x)+g(x)·f**^{ I}**(x)**Hence,

*h*

^{I}**(x) = f(x)·g**^{I}**(x)+g(x)·f**^{ I}**(x)**It's called the Product Rule of differentiation.

Different forms of notation of this rule are:

(1)

*[f(x)·g(x)]*

^{ I}**=**

= f= f

^{ I}**(x)·g(x)+f(x)·g**^{I}**(x)**(2)

*d*

=

**[f(x)·g(x)]**/dx ==

**f(x)·**d**g(x)**/dx+**g(x)·**d**f(x)**/dx*Examples*

*[x³·cos(x)]*

^{ I}**=**

= 3x²·cos(x)−x³·sin(x)= 3x²·cos(x)−x³·sin(x)

*d*

=

**(3x³·2**/dx =^{x})=

**3x³·ln(2)·2**^{x}+9x²·2^{x}*(d/dx)*

**[sin(2x)] =**

=(d/dx)=

**[2sin(x)·cos(x)] =**

= 2cos(x)·cos(x)−2sin(x)·sin(x)

= 2[cos²(x)−sin²(x)] =

= 2cos(2x)= 2cos(x)·cos(x)−2sin(x)·sin(x)

= 2[cos²(x)−sin²(x)] =

= 2cos(2x)

*D*

_{x}**[x³·5**

=D^{x}·cos(x)] ==

_{x}**[(x³·5**

=D^{x})·cos(x)] ==

_{x}**(x³·5**D^{x})·cos(x) + (x³·5^{x})·_{x}**[cos(x)] =**

=D=

_{x}**(x³)·5**D^{x}·cos(x) + x³·_{x}**(5**D^{x})·cos(x) + x³·5^{x}·_{x}**[cos(x)] =**

= 3x²·5= 3x²·5

^{x}·cos(x) + x³·ln(5)·5^{x}·cos(x) − x³·5^{x}·sin(x)
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