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Goten77

  • 3 years ago

hmm question is how to factor this below.. im doing something small and wrong

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  1. Goten77
    • 3 years ago
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    |dw:1353907783097:dw| so once i get the derivative and i factor out whatever how should it look?

  2. Goten77
    • 3 years ago
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    the 2nd term is to the power of 1/3

  3. Goten77
    • 3 years ago
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    hmm something like this |dw:1353908054340:dw|

  4. hartnn
    • 3 years ago
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    didn't u use product rule of differentiation ? know what is it ?

  5. Goten77
    • 3 years ago
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    i used product rule but kinda re arranged it

  6. hartnn
    • 3 years ago
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    \(y=(x+1)(x^2-1)^{1/3} \\ y'=(2x/3)(x+1)(x^2-1)^{-2/3}+(x^2-1) \\ y'=(x^2-1)[(2x/3)(x+1)(x^2-1)^{-2/3-1}+1]\) i think u had doubt in the last step i did .... ?

  7. Goten77
    • 3 years ago
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    the last step to get it into that simpler form... im not sure why i keep messing it up... its like when i set it = to 0 to find critical points i always do it right but for some reason when its to the power of 1/3 i miss up

  8. hartnn
    • 3 years ago
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    for x^2-1, i get x=1 or -1 for [...], there is no real solution....

  9. Goten77
    • 3 years ago
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    ye oh well i got the derivative right so eh hopefuly got some credit XD as for points i just put it in calculator and this wasnt the exactl problem but im sure i got some credit atleast thanks 4 help man

  10. hartnn
    • 3 years ago
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    welcome ^_^

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