Derivatives math 31

[kris.yarmak123]

Hello. I have a question, what do I do wrong that I cannot get the correct answer? My answer key says the answer is "(2√3)/3" and I got "√3". My solution and the question itself are attached below.

 

[pka]

I get \(y'=\dfrac{8x}{9(y-1)}\) so @\((3\sqrt{3},5)\) we have \(y'=\dfrac{24\sqrt{3}}{36}\).

 

[Dr.Peterson]

Hello. I have a question, what do I do wrong that I cannot get the correct answer? My answer key says the answer is "(2√3)/3" and I got "√3". My solution and the question itself are attached below.

All you did wrong was to cancel 16/18 incorrectly. It is not equal to 4/3, but 8/9. And 8/9 times (3√3)/4 is (2√3)/3.

 

[kris.yarmak123]

All you did wrong was to cancel 16/18 incorrectly. It is not equal to 4/3, but 8/9. And 8/9 times (3√3)/4 is (2√3)/3.

Oh, got it. Thank you! My carelessness is killing me ?

 

[Steven G]

16/18<1 while 4/3>1. You need to learn to catch that error.

 

[kris.yarmak123]

All you did wrong was to cancel 16/18 incorrectly. It is not equal to 4/3, but 8/9. And 8/9 times (3√3)/4 is (2√3)/3.

Thanks you very much!

 

[kris.yarmak123]

16/18<1 while 4/3>1. You need to learn to catch that error.

Thank you very much!

 

[kris.yarmak123]

I get \(y'=\dfrac{8x}{9(y-1)}\) so @\((3\sqrt{3},5)\) we have \(y'=\dfrac{24\sqrt{3}}{36}\).

Thank you very much!

 

[Steven G]

Oh, got it. Thank you! My carelessness is killing me ?

i know the feeling.