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

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====Example 3====
====Example 3====
Ex3:<math> f(x) = {{4x^4} \over {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.


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>

Revision as of 00:06, 25 August 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

Failed to parse (syntax error): {\displaystyle ]z'(v) =(v^3 + 5v)(4v^4)' - (4v^4)(v^3 + 5v)'\over (v^3 + 5v)^2 } }

Failed to parse (syntax error): {\displaystyle ]z'(v) =(v^3 + 5v)(16v^3) - (4v^4)(v^3 + 5v)'\over (v^3 + 5v)^2 } }