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IsTim Group Title

Express each function as a power with a rational exponent, and then differentiate. Express each answer using positive exponents. y=sqrt(2x-3x^5)

  • 2 years ago
  • 2 years ago

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  1. IsTim Group Title
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    Well. I'm just doing the first part. I got 1/2(2-15x^4)^-1/2

    • 2 years ago
  2. dumbcow Group Title
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    ok sqrt is same as exponent of 1/2 then apply chain rule to differentiate good..wait don't change inside...multiply whole thing by derivative of inside

    • 2 years ago
  3. dumbcow Group Title
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    \[(2-15x^{4})\frac{1}{2}(2x-5x^{3})^{-1/2}\]

    • 2 years ago
  4. dumbcow Group Title
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    **typo 3x^5

    • 2 years ago
  5. IsTim Group Title
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    In my textbook, it says: \[y=(2x-3x^5)^{1/2}\]

    • 2 years ago
  6. dumbcow Group Title
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    ?? yes that is original function right

    • 2 years ago
  7. IsTim Group Title
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    No, the answer. Well, the first half that isn't dy/dx.

    • 2 years ago
  8. dumbcow Group Title
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    oh yeah...sqrt is same as exponent of 1/2

    • 2 years ago
  9. IsTim Group Title
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    But I don't know how to get \[dy/dx=(2=15x^4 )/(2\sqrt{2x-3x^5}\]

    • 2 years ago
  10. IsTim Group Title
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    Also, why is their exponent positive, but ours is negative?

    • 2 years ago
  11. dumbcow Group Title
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    same answer i posted....ours is negative because its in numerator, when you flip it to denominator the sign of exponent changes

    • 2 years ago
  12. IsTim Group Title
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    Ok. I understand the first part. Thanks. I'll call out if I need help on the second part.

    • 2 years ago
  13. IsTim Group Title
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    Um, to find the derviative I use 1/2(2x−5x^3)−1/2 right.

    • 2 years ago
  14. dumbcow Group Title
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    whoah ...um we were doing 2nd part (finding derivative this whole time) or at least thats what i was doing

    • 2 years ago
  15. IsTim Group Title
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    Oh no! How do I do the first part?

    • 2 years ago
  16. dumbcow Group Title
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    sqrt is same as exponent of 1/2

    • 2 years ago
  17. IsTim Group Title
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    To get dy/dx=(2=15x^4)

    • 2 years ago
  18. IsTim Group Title
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    /(22x−3x5−−−−−−−√

    • 2 years ago
  19. IsTim Group Title
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    I guess I can't copy Latex....

    • 2 years ago
  20. IsTim Group Title
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    dy/dx=(2=15x^4)/(2sqrt(2x−3x^5))

    • 2 years ago
  21. IsTim Group Title
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    How do I figure that out?

    • 2 years ago
  22. dumbcow Group Title
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    let me try this, part 1 is super easy all you are doing is rewriting it using rational exponents \[\large \sqrt[n]{x} = x^{1/n}\] \[\sqrt{2x-3x^{5}} = (2x-3x^{5})^{1/2}\] part 2: find derivative using chain rule u = 2x-3x^5 du = 2-15x^4 \[\frac{d}{du} u^{1/2} = \frac{1}{2}u^{-1/2} = \frac{1}{2\sqrt{u}}\] multiply by du and substitute 2x-3x^5 back in for u \[\rightarrow \frac{2-15x^{4}}{2\sqrt{2x-3x^{5}}}\]

    • 2 years ago
  23. IsTim Group Title
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    How'd you get du = 2-15x^4?

    • 2 years ago
  24. IsTim Group Title
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    @dumbcow When you have time...

    • 2 years ago
  25. dumbcow Group Title
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    isn't that what you got in your very 1st post ^^ ?? anyway, its the derivative of 2x-3x^5 using power rule \[\large \frac{d}{dx} x^{n} = n*x^{n-1}\]

    • 2 years ago
  26. IsTim Group Title
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    Ok. thanks.

    • 2 years ago
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