Question

y'=? , y=x.(root of 8-x^2)

Answer 1

\[y=x \sqrt{8-x ^{2}} \] what is y' and y''

Answer 2

use the product rule,\[\frac{ d(g(x).f(x)) }{ dx }=g'(x)f(x)+g(x)f'(x)\]\[\times f(x)\]

Answer 3

ignore "x f(x)" at the bottom.

Answer 4

huh hahaha okey

Answer 5

what about the root

Answer 6

Use,\[ \frac{ d((f(x))^n) }{ dx}=n.f'(x).(f(x))^{ n-1 }\]

Answer 7

i stoped at \[\sqrt{8-x2}+x \frac{ -2x }{ 2\sqrt{8-x ^{2}} }=0\]

Answer 8

correct except it doesnt nessecarily =0, it is y'=

Answer 9

yeah

Answer 10

solved it x=0 ,thanks

Answer 11

Have you managed y''?