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sdavenko

  • 4 years ago

what is the derivative of x(2x-3)^1/2?

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  1. dpaInc
    • 4 years ago
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    (2x-3)^(1/2) + x(2x-3)^(-1/2)

  2. sdavenko
    • 4 years ago
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    what about the 2 that is taken from the inside of the function? and thats what i have already but i dont know how to simplify it further

  3. arcticf0x
    • 4 years ago
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    That 2 has been cancelled by 1/2 from the exponent. Do you have some answer you have to simplify for?

  4. sdavenko
    • 4 years ago
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    oh thanks butthis isnt an integration problem

  5. sdavenko
    • 4 years ago
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    the other options are a) 3x-3/ sqrt2x-3 b.) x/sqrt 2x-3 c.) 1/sqrt 2x-3 d.)-x+3/ sqrt 2x-3 e.)5x-6/2 sqrt 2x-3

  6. arcticf0x
    • 4 years ago
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    Simply add the terms, you will get 3x-3/(2x-3)^(1/2)...Know how?

  7. sdavenko
    • 4 years ago
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    no the final step im on is (2x-3)^1/2 +(2x)(1/2)(2x-3)^-1/2

  8. arcticf0x
    • 4 years ago
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    Ok, look, we have, (2x-3)^(1/2) +x(2x-3)^(-1/2) => [(2x-3)^(1/2) * (2x-3)^(1/2) + x ]/(2x-3)^(-1/2) =>(2x-3+x)/(2x-3)^(-1/2) =>3x-3/(2x-3)^(1/2)

  9. sdavenko
    • 4 years ago
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    ah i feel th feel like i have made this problem harder than it neeeds to be but, why did you multiply (2x-3)^(1/2) by (2x-3)^(1/2) in the second step?

  10. arcticf0x
    • 4 years ago
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    Its the way you add surds, multiply the denominator and join them.

  11. sdavenko
    • 4 years ago
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    ok thanks! I really appreciate you're patience

  12. arcticf0x
    • 4 years ago
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    I wish i could have done that in proper latex so it'd have been clear before.

  13. sdavenko
    • 4 years ago
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    where are you from? what country?

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