## ParthKohli 2 years ago More differentiation problems with quotient rule. I'm going to type out the question along with the answer. Thanks for the patience :)

1. ParthKohli

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

2. ParthKohli

$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.

3. ParthKohli

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

4. ParthKohli

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

5. ParthKohli

Can I simplify it more?

6. cwrw238

you can factorise the numerator i suppose

7. ParthKohli

Oh, yes. Is there anything else?

8. KingGeorge

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

9. ParthKohli

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

10. ParthKohli

Can we integrate that and check?

11. cwrw238

12. KingGeorge

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

13. ParthKohli

So then am I correct?

14. KingGeorge

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

15. ParthKohli

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

16. cwrw238

oh yes - just a typo

17. KingGeorge

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

18. ParthKohli

Thank you again!

19. cwrw238

well done parth - just a human error

20. ParthKohli

Thank you @cwrw238. Your encouragement much appreciated.

21. KingGeorge

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

22. ParthKohli

OK sir. Thank you for your help.