Practice Differentiate 1

harpazo

Pure Mathematics
Banned
Mar 20, 2018
5,788
361
83
NYC
Differentiate.

sDraw_2018-10-09_19-45-36.png
 

TheJason

Administrator
Staff member
Administrator
Moderator
Jan 25, 2018
2,508
545
113
jasonyankee.info
\(\displaystyle y = \sqrt[3]{x^{2}} * (2x - x^{2})\)

Without product rule:

\(\displaystyle y = x^{2/3}(2x - x^{2})\)

\(\displaystyle y = 2x^{5/3} - x^{8/3}\)

\(\displaystyle y' = \dfrac{5}{3}x^{2/3} - \dfrac{8}{3}x^{5/3}\)
 
Last edited:

MarkFL

La Villa Strangiato
Staff member
Administrator
Moderator
Math Helper
Jan 25, 2018
3,496
4,276
113
St. Augustine
Jason, the first term on the RHS of your final result has the original coefficient of 2 left off. :)
 

TheJason

Administrator
Staff member
Administrator
Moderator
Jan 25, 2018
2,508
545
113
jasonyankee.info
Jason, the first term on the RHS of your final result has the original coefficient of 2 left off. :)
\(\displaystyle y = \sqrt[3]{x^{2}} * (2x - x^{2})\)

Without product rule:

\(\displaystyle y = x^{2/3}(2x - x^{2})\)

\(\displaystyle y = 2x^{5/3} - x^{8/3}\)

\(\displaystyle y' = 2(\dfrac{5}{3})x^{2/3} - \dfrac{8}{3}x^{5/3}\)

\(\displaystyle y' = \dfrac{10}{3}x^{2/3} - \dfrac{8}{3}x^{5/3}\)
 
  • Like
Reactions: harpazo and MarkFL

harpazo

Pure Mathematics
Banned
Mar 20, 2018
5,788
361
83
NYC
Good job.