What is the simplified derivative of (50t)/(6+t)^3?

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SUMMARY

The simplified derivative of the function (50t)/(6+t)^3 can be accurately derived using the quotient rule. The correct answer is (-100(t - 3))/(6 + t)^4, as stated in the textbook. The common mistake involves overcomplicating the numerator after applying the quotient rule. Instead, factoring out common elements simplifies the expression effectively, leading to the correct result.

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Homework Statement



I keep getting it wrong. Using the quotient rule.

Homework Equations



Derivative of (50t)/(6+t)^3


The Attempt at a Solution



My answer: -100t^3 - 900t^2 + 10800/((6+t)^3)^2)
The book: (-100 (t - 3))/(6 + t)^4

What happened?



EDIT: Nevermind, I realize they're basically the same. I'll leave this here incase someone needs help in the future.
 
Last edited:
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how did you arrive at your answer? did you use the product rule?

with y=u v then dy/dx = du/dx .v + u . dv/dx

i get the book answer
 
939 said:

Homework Statement



I keep getting it wrong. Using the quotient rule.

Homework Equations



Derivative of (50t)/(6+t)^3


The Attempt at a Solution



My answer: -100t^3 - 900t^2 + 10800/((6+t)^3)^2)
The book: (-100 (t - 3))/(6 + t)^4

What happened?



EDIT: Nevermind, I realize they're basically the same. I'll leave this here incase someone needs help in the future.
To get the book's answer, don't multiply things out in the numerator after you use the quotient rule. In the numerator, both terms have a factor of 50 and a factor of (6 + t)2. Bring out these factors, and what you have left is easy to simplify, and results in the book's answer.
 

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