Difference between revisions of "Calculus:Derivative of Polynomial Functions"
Jump to navigation
Jump to search
(23 intermediate revisions by the same user not shown) | |||
Line 1: | Line 1: | ||
<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> | ||
Line 14: | Line 16: | ||
==Examples== | ==Examples== | ||
Find the derivative | |||
====Example 1==== | ====Example 1==== | ||
Ex1:<math>f | Ex1:<math>f(x^4+2x^2+4x+2)</math> | ||
<math>(f+g)'=f'+g'</math> | |||
Using the sum rule we can divided in to different part <math>f((x^4)+2(x^2)+4(x)+(2 | Using the sum rule we can divided in to different part <math>f'(x^4+2x^2+4x+2)=(x^4)+2(x^2)+4(x)+(2)</math> | ||
so we will started to work on different part by using power rule. | so we will started to work on different part by using power rule. | ||
<math>f'((x^4)+2(x^2)+4(x)+(2 | <math>f'(x)=(x^4)+2(x^2)+4(x)+(2)</math> | ||
<math>(4x^3)+2(2x)+4</math> | <math>(4x^3)+2(2x)+4</math> | ||
Line 28: | Line 33: | ||
====Example 2==== | ====Example 2==== | ||
Ex2:<math>f'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> | |||
Using the Product Rule we can divided in to different part <math>f'(x^4*x^3) = x^4 * (x^3)'+(x^4)' * x^3)</math> | |||
<math>x^4 * 3x^2 + 4x^3 * x^3</math> | |||
<math>3x^6 + 4x^6</math> | |||
<math>3x^6 + 4x^6</math> | |||
<math>7x^6</math> | |||
====Example 3==== | |||
Ex3:<math> z(v) = {{4v^4} \over {v^3 + 5v}} </math> | |||
Now we can understand v as x the idea will be the same. | |||
By using the quotient rule <math>({f \over g})' = {(gf'-fg') \over g^2} </math> | |||
we can under stand it as | |||
f(v)=4v^4 | |||
g(v)=v^3 + 5v | |||
so we will get | |||
<math> z'(v) ={(v^3 + 5v)(4v^4)' - (4v^4)(v^3 + 5v)' \over (v^3 + 5v)^2 } </math> | |||
= | <math> z'(v) ={ (v^3 + 5v)(16v^3) - (4v^4)(v^3 + 5v)' \over (v^3 + 5v)^2 } </math> | ||
</noinclude> | </noinclude> |
Latest revision as of 13:41, 16 September 2021
Derivative of Polynomial Functions
=Newton Derivative of Polynomial Functions=
- The sum rule
- The Difference Rule
- The Product Rule
- The Quotient Rule
=Leibniz Derivative of Polynomial Functions=
- The sum rule
- The Difference Rule
- The Product Rule
- 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