- #1

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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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- Thread starter Quadruple Bypass
- Start date

- #1

- 120

- 0

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?

- #2

TD

Homework Helper

- 1,022

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[tex]

\begin{gathered}

y = \left( {x^3 + y^3 } \right)^{20} \hfill \\

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 \\

\end{gathered}

[/tex]

Now you can solve for y'.

- #3

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

- #4

TD

Homework Helper

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Good luck!

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