Extended product rule for derivatives

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SUMMARY

The extended product rule for derivatives allows differentiation of functions with three or more terms. The standard product rule is expressed as (uv)' = uv' + u'v, where u and v are functions. For three functions, the derivative is given by (uvw)' = (uv)'w + (uv)w' = u'vw + uv'w + uvw'. This formula can be extended to four or more terms, facilitating the differentiation of complex products efficiently.

PREREQUISITES
  • Understanding of basic differentiation rules, specifically the product rule.
  • Familiarity with functions and their derivatives.
  • Knowledge of calculus concepts, particularly the chain rule.
  • Ability to manipulate algebraic expressions involving multiple functions.
NEXT STEPS
  • Study the derivation of the extended product rule using examples with three or more functions.
  • Practice differentiating complex functions using the extended product rule.
  • Explore applications of the extended product rule in real-world problems.
  • Learn about higher-order derivatives and their relationship with the product rule.
USEFUL FOR

Students of calculus, mathematics educators, and anyone looking to deepen their understanding of differentiation techniques, particularly those involving multiple functions.

musicfairy
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Can someone please explain it to me? My handwriting wasn't at its best when I was taking notes in class and now I can't read it. The teacher showed an example that I jotted down but what's the general rule?
 
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Maybe you should consult your textbook or take notes more carefully.
The usual product rule for differentiation reads
<br /> (uv)&#039;=uv&#039;+u&#039;v<br />
where u,v are functions.

What do you mean by extended product rule?
 
I meant if you have 3 or more terms, like y = sinxcosxlnx and you want the derivative.
 
For three terms: (uvw)'=(uv)'w+(uv)w'=u'vw+uv'w+uvw'.

As you can see, this can be extended to 4 terms and beyond quite readily.
 
It's easy to derive yourself, using the regular problem rule. If you have y = f(x)g(x)h(x), then use the formula Pete Callahan posted, using u=f(x), v=(g)h(x).
So then y' = uv' + vu'
Then use the product rule again to find v'
 
Thanks everyone. This makes more sense now.
 

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