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

Mathematics
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I'd need you guys to check it in a minute.
\[h(x) = {4\sqrt{x }\over x^2 - 2} \]\[h'(x) = {{(4\sqrt{x})'(x^2 - 2) - (4\sqrt{x})(x^2 - 2)'}\over (x^2 - 2)^2 } \]\[\implies\qquad{2x^{-1 \over 2} (x^2 - 2) - (4\sqrt{x})(2x) \over (x^2 - 2)^2} \]More steps coming.
\[\implies{(2x^{3 \over 2} - 4x^{-1 \over 2}) - 8x^{3 \over 2} \over (x^2 - 2)^2} \]

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Other answers:

\[\implies {{-6x^{3 \over 2}} - {4x^{1 \over 2} }\over(x^2 - 2)^2} \]
Can I simplify it more?
you can factorise the numerator i suppose
Oh, yes. Is there anything else?
Using wolfram, I'm not getting the same answer as that
Oh... I might be wrong in that case.
Can we integrate that and check?
i cant see any problem with your answer
\[\Large{(2x^{3 \over 2} - 4x^{-1 \over 2}) - 8x^{3 \over 2} \over (x^2 - 2)^2} \\ =(2x^{3 \over 2} -8x^{3 \over 2} ) - 4x^{-1 \over 2} \over (x^2 - 2)^2\\={{-6x^{3 \over 2}} - {4x^{-\frac{1}{2}} }\over(x^2 - 2)^2}\]
So then am I correct?
Your exponent on the 4 was positive when it was supposed to be negative.
Oh, I mistakenly made that positive when it was negative in the previous step. *close enough*
oh yes - just a typo
You can then factor the \[\frac{1}{\sqrt{x}}\]out, and make it look significantly simpler.
Thank you again!
well done parth - just a human error
Thank you @cwrw238. Your encouragement much appreciated.
\[\Large {{-6x^{3 \over 2}} - {4x^{-\frac{1}{2}} }\over(x^2 - 2)^2}\\ \Large =\frac{1}{\sqrt{x}}{{-6x^2} - {4 }\over(x^2 - 2)^2}\\ \Large =\frac{-6x^2-4}{\sqrt{x}(x^2-2)^2}\]Then if you want, you can factor a -2 out of the top as well, but this is as much as it simplifies.
OK sir. Thank you for your help.

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