Question

If \(x^2 + y^2 = 1^2\)

\(y' = -\)\(/\)

(for \(y \neq 0\) ) \(y'' = -\)\(/y^3\)

Answer 1

are you getting the question?

Answer 2

no

Answer 3

i have to find single and double derivative
of the given function

Answer 4

first derivative :2x+2y=0
second, 2+2=4

Answer 5

explain please

Answer 6

i know how to find the first derivative tell me about the second

Answer 7

ok
2x + 2y * dy/dx = 0
dy/dx = -2x / 2y = -x/y

Answer 8

what about the second derivative?

Answer 9

use the quotient rule

Answer 10

i tried but no success

Answer 11

\[y''=\frac{(-x)'y-(-x)y'}{y^2}\]

Answer 12

(-x)'=-1

Answer 13

second derivative use quotient rule

y * -1 - (-x) * dy/dx
-------------------
y^2

-y + x*dy/dx
----------
y^2

Answer 14

replace y' with -x/y

Answer 15

multiply top and bottom by y

Answer 16

you will be able to use x^2+y^2=1 to simplify further

Answer 17

got it thanks

Answer 18

thanks myin - i got a bit lost after the second line!

Answer 19

guys the answer im getting is wrong! :( can you solve it for me?

Answer 20

\[y''=\frac{(-x)'y-(-x)y'}{y^2}=\frac{-1y+xy'}{y^2}=\frac{-y+x \cdot \frac{-x}{y}}{y^2}\]
\[=\frac{-y \cdot y+x(-x)}{y \cdot y^2}=\frac{-y^2-x^2}{y^3}=\frac{-(x^2+y^2)}{y^3}\]

Answer 21

now remember x^2+y^2=1

Answer 22

\[=\frac{-1}{y^3}\]

Answer 23

very nice....

Answer 24

nice