Chain Rule for Derivatives: Differentiating a Product with Chain Rule

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Homework Help Overview

The discussion revolves around differentiating the function f(x)=(3x^{2}+4)^{3}(5-3x)^{4}, which involves both the product rule and the chain rule in calculus.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster attempts to clarify whether to evaluate the derivatives separately using the chain rule for each part of the product or if there is an alternative method for differentiation.

Discussion Status

Some participants have provided guidance on applying the chain rule in conjunction with the product rule. Others have suggested that expanding the functions could be an alternative approach, indicating that multiple methods are being considered.

Contextual Notes

There is no explicit consensus on the best approach, and the discussion includes varying interpretations of how to apply the rules of differentiation effectively.

mg0stisha
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Homework Statement


Differentiate [tex]f(x)=(3x^{2}+4)^{3}(5-3x)^{4}[/tex]



Homework Equations


N/A



The Attempt at a Solution


I can see that this derivative is a product, yet also involves using chain rule. With this being said, am i just supposed to evaluate these separately using chain rule for each then multiply the results together? Or is there another way to differentiate this? Thanks in advance.
 
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Take it as [tex]f(x) = g(u(x))h(v(x))[/tex]. Then [tex]f'(x) = g'(u)h(v) + h'(v)g(u)[/tex] where for example [tex]g'(u) = \frac{dg(u)}{dx} = \frac{dg(u)}{du}*\frac{du}{dx}[/tex], your standard chain rule.
 
Ah yes I see it now, thank you very much!
 
This is a product rule question, however, to take the derivative of this, you'll need the derivative of first and the derivative of the 2nd, thus the chain rule.

If you don't want to use the chain rule, you can expand both and use the product rule. :)
 

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