Mastering the Product Rule for Differentiation: A Comprehensive Guide

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Homework Help Overview

The discussion revolves around differentiating a product of two functions using the product rule in calculus, specifically the expression (2t^2+t^{(1/3)})(4t-5).

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of the product rule and the correctness of the differentiation steps. Questions arise regarding discrepancies between the original poster's result and the answer provided in a textbook.

Discussion Status

Some participants affirm the original poster's differentiation steps, while others point out that the answers may be equivalent despite appearing different. There is an exploration of how the terms can be matched and factored differently.

Contextual Notes

Participants note the presence of different variable representations (t vs. x) and question whether this affects the validity of the results. The original poster expresses confusion over the differences in outcomes.

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Homework Statement


Differentiate using the product rule

Homework Equations



(2t^2+t^{(1/3)})(4t-5)

The Attempt at a Solution



h'(t) = f'(t)g(t)+f(t)g'(t)

(4t+\frac {1}{3}t^{\frac {-2}{3}})(4t-5)+(2t^2+t^{(1/3)})(4)
\frac {-5}{3t^{2/3}}+24t^{2}-20t+\frac {16}{3}t^{1/3}

Why is this wrong?
 
Last edited:
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maybe your missing something in the question because the work you showed is correct.
 
Looks right as can be.
 
The answer in the book is much different, though:
h'(t) = \frac {72x^{8/3} - 60x^{5/3} + 16x -5}{3x^{2/3}}

Something is not right :/.
 
Those 2 answers are equal... other then the t's all of a sudden being x's :).
 
It's not that much different. 72*x^(8/3)/(3*x^(2/3)) is 24*x^2 which if I replace x by t corresponds to the 24*t^2 in your solution. Can you match the other terms up as well? They just factored the answer in a different way.
 
Ahhhhh, that explains it! Thank you very much :).
 

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