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ERoseM

  • 3 years ago

Find y prime if y=(x^2+3x-5)(cube root(x)-1)

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  1. nickhouraney
    • 3 years ago
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    i would probably start by rewriting it (x^2+3x-5)(x-1)^(1/3)

  2. ERoseM
    • 3 years ago
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    Okay should i do the chain rule?

  3. nickhouraney
    • 3 years ago
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    i did product rule

  4. nickhouraney
    • 3 years ago
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    u got it?

  5. ERoseM
    • 3 years ago
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    I got |dw:1350267174267:dw|

  6. nickhouraney
    • 3 years ago
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    dont think thats correct

  7. nickhouraney
    • 3 years ago
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    was the original problem (cuberoot(x)) - 1)

  8. nickhouraney
    • 3 years ago
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    only the x is under the cube root?

  9. ERoseM
    • 3 years ago
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    Only the x under the cube root that where I get stuck

  10. nickhouraney
    • 3 years ago
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    oh i did it \[\sqrt[3]{x-1}\] so you might be right

  11. ERoseM
    • 3 years ago
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    I just couldnt decide whether or not to do the chain rule and what to do with the (-1) if I do

  12. nickhouraney
    • 3 years ago
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    well this is a product rule (x^2+3x-5) d/dx(cube root(x)-1) + (cube root(x)-1) d/dx(x^2+3x-5)

  13. nickhouraney
    • 3 years ago
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    = (x^2+3x-5) 1/3 x^-2/3 + (x^1/3 - 1) (2x+3)

  14. nickhouraney
    • 3 years ago
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    just algebra and reducing after this point

  15. ERoseM
    • 3 years ago
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    Okay thank you!

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