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ParthKohli
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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 :)
 one year ago
 one year ago
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 :)
 one year ago
 one year ago

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ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
I'd need you guys to check it in a minute.
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
\[h(x) = {4\sqrt{x }\over x^2  2} \]\[h'(x) = {{(4\sqrt{x})'(x^2  2)  (4\sqrt{x})(x^2  2)'}\over (x^2  2)^2 } \]\[\implies\qquad{2x^{1 \over 2} (x^2  2)  (4\sqrt{x})(2x) \over (x^2  2)^2} \]More steps coming.
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
\[\implies{(2x^{3 \over 2}  4x^{1 \over 2})  8x^{3 \over 2} \over (x^2  2)^2} \]
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
\[\implies {{6x^{3 \over 2}}  {4x^{1 \over 2} }\over(x^2  2)^2} \]
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
Can I simplify it more?
 one year ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.1
you can factorise the numerator i suppose
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
Oh, yes. Is there anything else?
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
Using wolfram, I'm not getting the same answer as that
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
Oh... I might be wrong in that case.
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
Can we integrate that and check?
 one year ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.1
i cant see any problem with your answer
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
\[\Large{(2x^{3 \over 2}  4x^{1 \over 2})  8x^{3 \over 2} \over (x^2  2)^2} \\ =(2x^{3 \over 2} 8x^{3 \over 2} )  4x^{1 \over 2} \over (x^2  2)^2\\={{6x^{3 \over 2}}  {4x^{\frac{1}{2}} }\over(x^2  2)^2}\]
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
So then am I correct?
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
Your exponent on the 4 was positive when it was supposed to be negative.
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
Oh, I mistakenly made that positive when it was negative in the previous step. *close enough*
 one year ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.1
oh yes  just a typo
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
You can then factor the \[\frac{1}{\sqrt{x}}\]out, and make it look significantly simpler.
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
Thank you again!
 one year ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.1
well done parth  just a human error
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
Thank you @cwrw238. Your encouragement much appreciated.
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
\[\Large {{6x^{3 \over 2}}  {4x^{\frac{1}{2}} }\over(x^2  2)^2}\\ \Large =\frac{1}{\sqrt{x}}{{6x^2}  {4 }\over(x^2  2)^2}\\ \Large =\frac{6x^24}{\sqrt{x}(x^22)^2}\]Then if you want, you can factor a 2 out of the top as well, but this is as much as it simplifies.
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
OK sir. Thank you for your help.
 one year ago
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