Question

need help finding the second derivative of
sinx+x^2y=10

Answer 1

\[sinx+x^2y=10\]

Answer 2

\[\cos(x)+x^2y'+2xy=0\]
\[ x^2y'=-2xy-\cos(x)\]
\[y'=\frac{-2xy-\cos(x)}{x^2}\]

Answer 3

now apply quotient rule to find y''

Answer 4

ok need a minute to draw it out.
How do you get the fraction bar to work in equation editor?

Answer 5

frac{ }{ }

Answer 6

ok let me retype it again

Answer 7

\[y''=\left[ -2x^2y-2x^3y'+x^2\sin(x)-4x^2y+2xcos(x) \right]/x^4\]

Answer 8

\[y''=\frac{[-2(xy'+y)+\sin(x)]x^2-2x[-2xy-\cos(x)]}{x^4}\]

Answer 9

i had to do it myself
this one is a little more uglier

Answer 10

yes it is. I wonder if I can put these into wolfram alpha?

Answer 11

\[y''=\frac{-2x^3y'-2x^2y+x^2\sin(x)+4x^2y+2x \cos(x)}{x^4}\]

Answer 12

yes i think you can
i will look at it in a second and see okay?

Answer 13

ok I see I made a sign mistake

Answer 14

\[y''=\frac{-2x^3 (\frac{-2xy-\cos(x)}{x^2})-2x^2y+x^2 \sin(x)+4x^2y+2x \cos(x)}{x^4}\]

Answer 15

are you using frac{}|{} in the equation editor ? I tried and it did not seem to work

Answer 16

yes don't put that | thing in between the {} {}

Answer 17

just put frac{ }{ }

Answer 18

numerator inside first one
denominator inside second one

Answer 19

\[y'=\frac{2}{3}\]

Answer 20

cool thanks, I just wrote any fraction to play around

Answer 21

thanks for your help. I was playing around in wolframalpha and was not able to do the second derivatives.

Answer 22

i'm still trying

Answer 23

lol i don't know if i can figure it out you know how to use wofram to do it

Answer 24

i can only get it to tell me the first derivative sorry

Answer 25

myininaya
can you tell me if
the second derivative of
\[2x^2+y^2=4\]
is\[\frac{d^2y}{dx}=\frac{-8}{y^3}\]

Answer 26

Thanks in advance for all of your help tonight :)

Answer 27

\[4x+2yy'=0 => y'=\frac{-4x}{2y}=\frac{-2x}{y} => y''=\frac{-2y-(-2x)y'}{y^2}\]
\[y''=\frac{-2y+2x \frac{-2x}{y}}{y^2}=\frac{-2y+-2x^2 \cdot \frac{1}{y}}{y^2} \cdot \frac{y}{y}\]
\[y''=\frac{-2y^2-2x^2}{y^3}=\frac{-2(x^2+y^2)}{y^3}=\]
.... i think i did something wrong

Answer 28

oh i see it

Answer 29

I got \[\frac{d^2y}{dx}=\frac{-4x^2-2y^2}{y^3}\]

Answer 30

\[y''=\frac{-2y+2x \frac{-2x}{y}}{y^2}=\frac{-2y+-4x^2 \cdot \frac{1}{y}}{y^2} \cdot \frac{y}{y}\]
\[y''=\frac{-2y^2-4x^2}{y^3}=\frac{-2(2x^2+y^2)}{y^3}=\]

Answer 31

\[=\frac{-2(4)}{y^3}=\frac{-8}{y^3}\]

Answer 32

Thanks you are awesome. I enjoy your company and help :)

Answer 33

:)

Answer 34

sinx+x^2 y=10
cosx + 2xy + x^2y' = 0
-> y' = (- cosx + 2xy ) / x^2

-> y" = x^2 [ sinx + 2( y + xy') ] - 2x (- cosx + 2xy ) / x^4
=> y" = x^2sinx - 2x^2y + 2xcosx + 2x^3y'
=> y" = ( xsinx - 2xy +2cosx + 2x^2y') / x^3