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IsTim

  • 2 years ago

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

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

  2. dumbcow
    • 2 years ago
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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

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

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

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

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

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

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

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

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

  11. dumbcow
    • 2 years ago
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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

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

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

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

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

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

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

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

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

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

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

  22. dumbcow
    • 2 years ago
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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 \[\rightarrow \frac{2-15x^{4}}{2\sqrt{2x-3x^{5}}}\]

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

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

  25. dumbcow
    • 2 years ago
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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 \[\large \frac{d}{dx} x^{n} = n*x^{n-1}\]

  26. IsTim
    • 2 years ago
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    Ok. thanks.

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