Question

More differentiation problems with quotient rule. I'm going to type out the question along with the answer.

Thanks for the patience :)

Answer 1

I'd need you guys to check it in a minute.

Answer 2

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

Answer 3

\[\implies{(2x^{3 \over 2} - 4x^{-1 \over 2}) - 8x^{3 \over 2} \over (x^2 - 2)^2} \]

Answer 4

\[\implies {{-6x^{3 \over 2}} - {4x^{1 \over 2} }\over(x^2 - 2)^2} \]

Answer 5

Can I simplify it more?

Answer 6

you can factorise the numerator i suppose

Answer 7

Oh, yes. Is there anything else?

Answer 8

Using wolfram, I'm not getting the same answer as that

Answer 9

Oh... I might be wrong in that case.

Answer 10

Can we integrate that and check?

Answer 11

i cant see any problem with your answer

Answer 12

\[\Large{(2x^{3 \over 2} - 4x^{-1 \over 2}) - 8x^{3 \over 2} \over (x^2 - 2)^2} \\
=(2x^{3 \over 2} -8x^{3 \over 2} ) - 4x^{-1 \over 2} \over (x^2 - 2)^2\\={{-6x^{3 \over 2}} - {4x^{-\frac{1}{2}} }\over(x^2 - 2)^2}\]

Answer 13

So then am I correct?

Answer 14

Your exponent on the 4 was positive when it was supposed to be negative.

Answer 15

Oh, I mistakenly made that positive when it was negative in the previous step.
*close enough*

Answer 16

oh yes - just a typo

Answer 17

You can then factor the \[\frac{1}{\sqrt{x}}\]out, and make it look significantly simpler.

Answer 18

Thank you again!

Answer 19

well done parth - just a human error

Answer 20

Thank you @cwrw238. Your encouragement much appreciated.

Answer 21

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

Answer 22

OK sir. Thank you for your help.