Question

how do u work this out √(y^15)

Answer 1

y^(15/2) e^(i pi floor(1/2-(15 arg(y))/(2 pi)))

Answer 2

y^(15/2)

Answer 3

7 remainer 1

Answer 4

y^(15/2)+O(y^(25/2))

Answer 5

y^(15/2)+O((1/y)^(25/2))

Answer 6

d/dy(sqrt(y^15)) = (15 sqrt(y^15))/(2 y)

Answer 7

where did u get 25

Answer 8

Possible derivation:
d/dy(sqrt(y^15))
| Use the chain rule, d/dy(sqrt(y^15)) = ( du)/( du) ( du)/( dy), where u = sqrt(y^15) and ( du)/( du) = HoldForm'(u):
= | HoldForm'(sqrt(y^15)) (d/dy(sqrt(y^15)))
| Use the chain rule, d/dy(sqrt(y^15)) = ( dsqrt(u))/( du) ( du)/( dy), where u = y^15 and ( dsqrt(u))/( du) = 1/(2 sqrt(u)):
= | HoldForm'(sqrt(y^15)) (d/dy(y^15))/(2 sqrt(y^15))
| The derivative of y^15 is 15 y^14:
= | ((15 y^14) HoldForm'(sqrt(y^15)))/(2 sqrt(y^15))

Answer 9

y = 0

Answer 10

it say

Answer 11

@best.shakir are you copy-pasting from wolframalpha?

Answer 12

simplyfy √(y^15)

Answer 13

y sq root 15

Answer 14

see here
http://www.wolframalpha.com/input/?i=root+of+y%5E15

Answer 15

yes best shakir l dnt understand

Answer 16

\[\huge \sqrt[m]{x^n}\]
does that help @tafara27 ?

Answer 17

@lgbasallote u are good at explain

Answer 18

uhh wait...wrote that wrong

Answer 19

\[\huge \sqrt[m]{x^n} \implies x^{n/m}\]
does that help?

Answer 20

yea thnks

Answer 21

welcome