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anonymous

  • 4 years ago

Let y =(root5(x^8+5)). Find values A and B so that (dy)/(dx) = A(x^8+5)^B(C x^D)

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  1. myininaya
    • 4 years ago
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    \[y=\sqrt[5]{x^8+5}\] \[y=(x^8+5)^\frac{1}{5}\] Use the chain rule to find y'

  2. anonymous
    • 4 years ago
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    what if I need to find A,B,C, and D? what do I plug in?

  3. myininaya
    • 4 years ago
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    You need to find y' first to find A,B,C,and D

  4. anonymous
    • 4 years ago
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    I havent learned the chain rule yet

  5. myininaya
    • 4 years ago
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    \[[f(g(x))]'=f'(g(x)) \cdot g'(x) \text{ chain rule }\]

  6. myininaya
    • 4 years ago
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    Well you need the chain rule to do this problem.

  7. myininaya
    • 4 years ago
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    what is g(x) and what is f(x) if \[f(g(x))=(x^8+5)^\frac{1}{5}\]

  8. anonymous
    • 4 years ago
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    g(x)= 8x f(x)= 5^1/5

  9. myininaya
    • 4 years ago
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    No g(x) is the inside function and f(x) is the outside function

  10. anonymous
    • 4 years ago
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    oh ok g(x)= 8x^7

  11. myininaya
    • 4 years ago
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    \[g(x)=x^8+5 ; f(x)=x^\frac{1}{5}\] \[g'(x)=8x^7+0 ; f(x)=\frac{1}{5}x^{\frac{1}{5}-1}=\frac{1}{5}x^\frac{-4}{5}\]

  12. anonymous
    • 4 years ago
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    so I plug this in in the equation now

  13. myininaya
    • 4 years ago
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    this one \[y'=[f(g(x))]'=f'(g(x)) \cdot g'(x) \text{ chain rule } \] yes

  14. anonymous
    • 4 years ago
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    1/5x^-4/5

  15. myininaya
    • 4 years ago
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    \[y'=f'(x^8+5) \cdot 8x^7\]

  16. anonymous
    • 4 years ago
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    now just multiply, right?

  17. myininaya
    • 4 years ago
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    you need to plug x^8+5 into f'

  18. anonymous
    • 4 years ago
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    oh ok

  19. myininaya
    • 4 years ago
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    \[g'(x)=8x^7+0 ; f'(x)=\frac{1}{5}x^{\frac{1}{5}-1}=\frac{1}{5}x^\frac{-4}{5}\]

  20. myininaya
    • 4 years ago
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    f' is that second one I wrote down just now

  21. myininaya
    • 4 years ago
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    \[y'=f'(x^8+5) \cdot 8x^7=\frac{1}{5}(x^8+5)^\frac{-4}{5} \cdot 8x^7\]

  22. myininaya
    • 4 years ago
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    now you should be able to put it in the form above to get your A,B,C, and D

  23. anonymous
    • 4 years ago
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    i'll try it

  24. myininaya
    • 4 years ago
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    actually it is already in that form

  25. anonymous
    • 4 years ago
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    so just look for the A,B,C,D then right?

  26. myininaya
    • 4 years ago
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    yes

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