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ParthKohli Group Title

More differentiation problems with quotient rule. I'm going to type out the question along with the answer. Thanks for the patience :)

  • 2 years ago
  • 2 years ago

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  1. ParthKohli Group Title
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    I'd need you guys to check it in a minute.

    • 2 years ago
  2. ParthKohli Group Title
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    \[h(x) = {4\sqrt{x }\over x^2 - 2} \]\[h'(x) = {{(4\sqrt{x})'(x^2 - 2) - (4\sqrt{x})(x^2 - 2)'}\over (x^2 - 2)^2 } \]\[\implies\qquad{2x^{-1 \over 2} (x^2 - 2) - (4\sqrt{x})(2x) \over (x^2 - 2)^2} \]More steps coming.

    • 2 years ago
  3. ParthKohli Group Title
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    \[\implies{(2x^{3 \over 2} - 4x^{-1 \over 2}) - 8x^{3 \over 2} \over (x^2 - 2)^2} \]

    • 2 years ago
  4. ParthKohli Group Title
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    \[\implies {{-6x^{3 \over 2}} - {4x^{1 \over 2} }\over(x^2 - 2)^2} \]

    • 2 years ago
  5. ParthKohli Group Title
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    Can I simplify it more?

    • 2 years ago
  6. cwrw238 Group Title
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    you can factorise the numerator i suppose

    • 2 years ago
  7. ParthKohli Group Title
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    Oh, yes. Is there anything else?

    • 2 years ago
  8. KingGeorge Group Title
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    Using wolfram, I'm not getting the same answer as that

    • 2 years ago
  9. ParthKohli Group Title
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    Oh... I might be wrong in that case.

    • 2 years ago
  10. ParthKohli Group Title
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    Can we integrate that and check?

    • 2 years ago
  11. cwrw238 Group Title
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    i cant see any problem with your answer

    • 2 years ago
  12. KingGeorge Group Title
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    \[\Large{(2x^{3 \over 2} - 4x^{-1 \over 2}) - 8x^{3 \over 2} \over (x^2 - 2)^2} \\ =(2x^{3 \over 2} -8x^{3 \over 2} ) - 4x^{-1 \over 2} \over (x^2 - 2)^2\\={{-6x^{3 \over 2}} - {4x^{-\frac{1}{2}} }\over(x^2 - 2)^2}\]

    • 2 years ago
  13. ParthKohli Group Title
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    So then am I correct?

    • 2 years ago
  14. KingGeorge Group Title
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    Your exponent on the 4 was positive when it was supposed to be negative.

    • 2 years ago
  15. ParthKohli Group Title
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    Oh, I mistakenly made that positive when it was negative in the previous step. *close enough*

    • 2 years ago
  16. cwrw238 Group Title
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    oh yes - just a typo

    • 2 years ago
  17. KingGeorge Group Title
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    You can then factor the \[\frac{1}{\sqrt{x}}\]out, and make it look significantly simpler.

    • 2 years ago
  18. ParthKohli Group Title
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    Thank you again!

    • 2 years ago
  19. cwrw238 Group Title
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    well done parth - just a human error

    • 2 years ago
  20. ParthKohli Group Title
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    Thank you @cwrw238. Your encouragement much appreciated.

    • 2 years ago
  21. KingGeorge Group Title
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    \[\Large {{-6x^{3 \over 2}} - {4x^{-\frac{1}{2}} }\over(x^2 - 2)^2}\\ \Large =\frac{1}{\sqrt{x}}{{-6x^2} - {4 }\over(x^2 - 2)^2}\\ \Large =\frac{-6x^2-4}{\sqrt{x}(x^2-2)^2}\]Then if you want, you can factor a -2 out of the top as well, but this is as much as it simplifies.

    • 2 years ago
  22. ParthKohli Group Title
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    OK sir. Thank you for your help.

    • 2 years ago
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