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

__Derivative Properties -__

Compound Functions

Compound Functions

Our purpose is to express the derivative of a compound function (a function of a function) in terms of derivatives of its components.

Assume that the real function

**is defined and**

*f(x)**differentiable*(that is, its derivative exist) on certain interval with a derivative

*f**.*

^{ I}**(x)**Assume further that the real function

**is defined and**

*g(x)**differentiable*on certain interval with a derivative

*g**and its values are falling within the domain of function*

^{ I}**(x)****.**

*f(x)*Then we can talk about a

*compound*function

**.**

*h(x)=f(g(x))*In other words, this compound function can be represented as

**, where**

*h(x)=f(y)***.**

*y=g(x)*Let's determine the derivative of this compound function.

The increment of function

**is**

*h(x)=f(g(x))*Δ

**=**

*h(x)***Δ**

*h(x+***Δ**

*x)−h(x) =*

= f(g(x+= f(g(x+

*x)) − f(g(x))*Using the definition of the increment of a function

**Δ**

*g(x+***Δ**

*x) = g(x) +***,**

*g(x)*it would look like

Δ

**=**

*h(x)***Δ**

*h(x+***Δ**

*x)−h(x) =*

= f(g(x)+= f(g(x)+

*g(x)) − f(g(x))*Recalling the representation

**, we can write this as**

*y=g(x)*Δ

**Δ**

*h(x) = f(y+*

*y) − f(y)*where

**.**

*y=g(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(y+= [f(y+

**Δ**

*y) − f(y)]/***Δ**

*x =*

==

**Δ**

*f(y)/*

*x*where

**.**

*y=g(x)*To transform this into expressions that would lead us to separation of functions combined in this compound function, multiply and divide this by Δ

**.**

*y*Δ

**Δ**

*h(x)/*

*x =*= [Δ

**Δ**

*f(y)/***]·[Δ**

*y***/Δ**

*y***]**

*x*Notice that if Δ

**→**

*x***then Δ**

*0***→**

*y***since function**

*0***is differentiable and, since Δ**

*y=g(x)***→**

*y***, Δ**

*0***→**

*f(y)***since function**

*0***is differentiable.**

*f(x)*Therefore, if Δ

**→**

*x***,**

*0*[Δ

**Δ**

*f(y)/***] →**

*y**d*

**f(y)**/d**y**where

**and**

*y=g(x)*[Δ

**Δ**

*y/***] →**

*x**d*

**y**/d**x=**d**g(x)**/d**x**Finally, we came to an expression for a derivative of a compound function.

Δ

**Δ**

*h(x)/***→ [**

*x**d*]·[

**f(y)**/d**y***d*]

**y**/d**x**In other words,

*d*d

**h(x)/****= [**

*x**d*]·[

**f(y)**/d**y***d*]

**y**/d**x**where

**.**

*y=g(x)*In terms of original functions,

*d*d

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

*x*= [

*d*]·[

**f(y)**/d**y***d*]

**g(x)**/d**x**where

**.**

*y=g(x)*This is a major formula of differentiation of a compound function.

*Example 1*

*f(x)=cos(x); g(x)=x²*

*f(g(x))=cos(x²)**d*

=[

**cos(x²)/**d**=***x*=

*d*]·[

**cos(y)**/d**y***d*] =

**(x²)**/d**x**= [

**]·[**

*−sin(y)***]**

*2x*where

**.**

*y=x²*Therefore,

*d*d

**cos(x²)/****= −2·x·sin(x²)**

*x**Example 2*

*f(x)=1/x(x); g(x)=cos(x)*

*f(g(x))=1/cos(x)=sec(x)*[

*]*

**sec(x)***=*

^{I}=[

**]**

*1/cos(x)**=*

^{I}=[

**]**

*1/y**·[*

^{I}**]**

*cos(x)**=*

^{I}=[

**]·[**

*−1/y²***]**

*−sin(x)*where

**.**

*y=cos(x)*Therefore,

[

**]**

*sec(x)**=*

^{I}

*sin(x)/cos²(x)**Example 3*

*f(x)=e*^{x}; g(x)=sin(x)

*f(g(x))=e*^{sin(x)}[

*]*

**e**^{sin(x)}*=*

^{I}=[

**]**

*e*^{y}*·[*

^{I}**]**

*sin(x)**=*

^{I}=[

**]·[**

*e*^{y}**]**

*cos(x)*where

**.**

*y=sin(x)*Therefore,

[

**]**

*e*^{sin(x)}*=*

^{I}**·**

*e*^{sin(x)}

*cos(x)**Example 4*

*f(x)=x²; g(x)=sin(x)*

*f(g(x))=(sin(x))²=sin²(x)*[

*]*

**sin²(x)***=*

^{I}=[

**]**

*y²**·[*

^{I}**]**

*sin(x)**=*

^{I}=[

**]·[**

*2y***]**

*cos(x)*where

**.**

*y=sin(x)*Therefore,

[

**]**

*sin²(x)**=*

^{I}

*2·sin(x)·cos(x)*
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