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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 :)
 2 years ago
 2 years 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 :)
 2 years ago
 2 years 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.
 2 years 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.
 2 years ago

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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