## Goten77 2 years ago hmm question is how to factor this below.. im doing something small and wrong

1. Goten77

|dw:1353907783097:dw| so once i get the derivative and i factor out whatever how should it look?

2. Goten77

the 2nd term is to the power of 1/3

3. Goten77

hmm something like this |dw:1353908054340:dw|

4. hartnn

didn't u use product rule of differentiation ? know what is it ?

5. Goten77

i used product rule but kinda re arranged it

6. hartnn

\(y=(x+1)(x^2-1)^{1/3} \\ y'=(2x/3)(x+1)(x^2-1)^{-2/3}+(x^2-1) \\ y'=(x^2-1)[(2x/3)(x+1)(x^2-1)^{-2/3-1}+1]\) i think u had doubt in the last step i did .... ?

7. Goten77

the last step to get it into that simpler form... im not sure why i keep messing it up... its like when i set it = to 0 to find critical points i always do it right but for some reason when its to the power of 1/3 i miss up

8. hartnn

for x^2-1, i get x=1 or -1 for [...], there is no real solution....

9. Goten77

ye oh well i got the derivative right so eh hopefuly got some credit XD as for points i just put it in calculator and this wasnt the exactl problem but im sure i got some credit atleast thanks 4 help man

10. hartnn

welcome ^_^