Finding the Derivative of a Function with Implicit Differentiation

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Homework Help Overview

The discussion revolves around finding the derivative of a function using implicit differentiation, specifically for the expression (x^3+y^3)^20, where y is considered a function of x.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to find the derivative and presents their result, questioning its correctness. Another participant points out the need to consider y as an unknown function of x and provides a revised approach to differentiate the expression.

Discussion Status

The discussion is active, with participants engaging in clarifying the differentiation process. Guidance has been offered regarding the correct application of implicit differentiation, and the original poster expresses gratitude for the assistance received.

Contextual Notes

The original poster mentions a grading context, indicating that their initial attempt was partially correct but resulted in a minor deduction of points. There is an implied urgency due to an upcoming final exam.

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Assume that y is a function of x . Find y' = dy/dx for (x^3+y^3)^20

when i solved this i got y'= (20(x^3+y^3)^19 * 3x^2)/(-3y^2)

is this correct or am i missing something?
 
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It's not entirely right, remember that y(x) is an unknown function of x!

[tex] \begin{gathered}<br /> y = \left( {x^3 + y^3 } \right)^{20} \hfill \\<br /> y' = 20\left( {x^3 + y^3 } \right)^{19} \cdot \left( {x^3 + y^3 } \right)^\prime = 20\left( {x^3 + y^3 } \right)^{19} \cdot \left( {3x^2 + 3y^2 \cdot y'} \right) \hfill \\ <br /> \end{gathered} [/tex]

Now you can solve for y'.
 
thanks a lot man. the grader only took off 3 pts for that prob and didnt say anything else, so i didnt know what i did wrong.

THANKS A LOT, you just saved me from making several mistakes on my final exam tommorow :D
 
Good luck! :smile:
 

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