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> | ||
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====Example 3==== | ====Example 3==== | ||
Ex3:<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. | |||
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