Implicit Differentiation

  • #1
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?
 

Answers and Replies

  • #2
TD
Homework Helper
1,022
0
It's not entirely right, remember that y(x) is an unknown function of x!

[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
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
 
  • #4
TD
Homework Helper
1,022
0
Good luck! :smile:
 

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