Difference between revisions of "Calculus:Derivative of Polynomial Functions"

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<noinclude>
===[[Calculus:Derivative of Polynomial Functions|Derivative of Polynomial Functions]]===
===[[Calculus:Derivative of Polynomial Functions|Derivative of Polynomial Functions]]===
</noinclude>
=======[[use Notation::Newton]] Derivative of Polynomial Functions=======
=======[[use Notation::Newton]] Derivative of Polynomial Functions=======
#The sum rule <math>(f+g)'=f'+g'</math>
#The sum rule <math>(f+g)'=f'+g'</math>
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==Examples==
==Examples==
Find the derivative
====Example 1====
====Example 1====
Ex1:<math>f'(x^4+2x^2+4x+2)</math>
Ex1:<math>f(x^4+2x^2+4x+2)</math>


<math>(f+g)'=f'+g'</math>
<math>(f+g)'=f'+g'</math>
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====Example 2====
====Example 2====
Ex2:<math>f'(x)=x^4*x^3</math>
Ex2:<math>f(x)=x^4*x^3</math>


The Product Rule <math>(f*g)'=f*g'+ g*f'</math>
The Product Rule <math>(f*g)'=f*g'+ g*f'</math>
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====Example 3====
====Example 3====
Ex3:<math> f'(x) = {{4x^4 - x^2 + 10x} \over {4x^5 - x^3 + 5x}} </math>
Ex3:<math> z(v) = {{4v^4} \over {v^3 + 5v}} </math>
 
Now we can understand v as x the idea will be the same.


The Quotient Rule <math>({f \over g})' = {(gf'-fg') \over g^2} </math>
By using the quotient rule <math>({f \over g})' = {(gf'-fg') \over g^2} </math>


<math> f'(x) = {{(4x^5 - x^3 + 5x)}{(4x^4 - x^2 + 10x)'}-(4x^4 - x^2 + 10x){(4x^5 - x^3 + 5x)'} \over {(4x^5 - x^3 + 5x)}^2} </math>
we can under stand it as


<math> f'(x) = {{(4x^5 - x^3 + 5x)}{(4x^4 - x^2 + 10x)'}-(4x^4 - x^2 + 10x){(4x^5 - x^3 + 5x)'} \over {(16x^8-8x^6+80x^5+x^4-20x^3+100x^2)}} </math>
f(v)=4v^4
g(v)=v^3 + 5v


<math> f'(x) = {{(4x^5 - x^3 + 5x)}{(16x^3 - 2x + 10)}-(4x^4 - x^2 + 10x){(4x^5 - x^3 + 5x)'} \over {(16x^8-8x^6+80x^5+x^4-20x^3+100x^2)}} </math>
so we will get


<math> f'(x) = {{(4x^5 - x^3 + 5x)}{(16x^3 - 2x + 10)}-(4x^4 - x^2 + 10x){(20x^4 - 3x^2 + 5)} \over {(16x^8-8x^6+80x^5+x^4-20x^3+100x^2)}} </math>
<math> z'(v) ={(v^3 + 5v)(4v^4)' - (4v^4)(v^3 + 5v)' \over (v^3 + 5v)^2 } </math>


<math> f'(x) = {{(4x^5 - x^3 + 5x)}{(16x^3 - 2x + 10)}-(4x^4 - x^2 + 10x){(20x^4 - 3x^2 + 5)} \over {(16x^8-8x^6+80x^5+x^4-20x^3+100x^2)}} </math>
<math> z'(v) ={ (v^3 + 5v)(16v^3) - (4v^4)(v^3 + 5v)' \over (v^3 + 5v)^2 } </math>
<math>f'(x) ={ {(64x^8 - 24x^6 + 40x^5 + 82x^4 - 10x^3 - 10x^2 + 50x) - (80x^8 - 32x^6 + 200x^5 + 23x^4 - 30x^3 - 5x^2 + 50x)} \over {(16x^8-8x^6+80x^5+x^4-20x^3+100x^2)}}</math>
</noinclude>
</noinclude>

Latest revision as of 13:41, 16 September 2021

Derivative of Polynomial Functions

=Newton Derivative of Polynomial Functions=
  1. The sum rule
  2. The Difference Rule
  3. The Product Rule
  4. The Quotient Rule
=Leibniz Derivative of Polynomial Functions=
  1. The sum rule
  2. The Difference Rule
  3. The Product Rule
  4. The Quotient Rule


Examples

Find the derivative

Example 1

Ex1:

Using the sum rule we can divided in to different part

so we will started to work on different part by using power rule.

Example 2

Ex2:

The Product Rule

Using the Product Rule we can divided in to different part

Example 3

Ex3:

Now we can understand v as x the idea will be the same.

By using the quotient rule

we can under stand it as

f(v)=4v^4 g(v)=v^3 + 5v

so we will get