Question

If 3x^2 + y^2 = 7 then evaluate the second derivative of y with respect to x when x = 1 and y = 2. Round your answer to 2 decimal places. Use the hyphen symbol, -, for negative values.

Answer 1

On this, I got the first derivative rather easily, but am having difficulty finding the second derivative.

Answer 2

The first derivative being -3x/y

Answer 3

@freckles

Answer 4

y'=-3x/y is what I meant.

Answer 5

Try for second

Answer 6

I did try for the second, I just get stuck in a mess.

Answer 7

Its a u/v type function you know how to differentiate them

Answer 8

Or simply
1st = 6x+2y(dy/dx)=0

Answer 9

\[\Large \frac{dy}{dx} = -\frac{3x}{y}\] is the correct first derivative

Answer 10

Here's my work for the second derivative, and tell me where I went wrong

Answer 11

Isn't that what I have?

Answer 12

So 2nd= 6 + 2(dy/dx)^2 +2y (y'')

Answer 13

yes @Tahamohammed I was confirming you had the right answer

Answer 14

\[\Large \frac{dy}{dx} = -\frac{3x}{y}\]

\[\Large \frac{d}{dx}\left[\frac{dy}{dx}\right] = \frac{d}{dx}\left[-\frac{3x}{y}\right]\]

use the quotient rule

Answer 15

6x+2yy'
6+2y'^2+2yy"
y"=6+2y'^2/-2y
y"=(6+2(-3x/y)^2)/-2y
y"=(6+18x^2/y^2)/-2y
y"=(6y^2/y^2+18x^2/y^2)/-2y

Answer 16

Where dy/DX=-3(1)/2=-3/2
So 6+2(-3/2)^2+4(y'')=0
You have to find y''

Answer 17

Did I make any mistakes so far? @Pawanyadav @jim_thompson5910 ?

Answer 18

Now just plug values you are correct

Answer 19

Just plug in the values?

Answer 20

Shouldn't I simplify further?

Answer 21

No need of it ......but if you want to then you can do

Answer 22

So is the answer -2.625 then?

Answer 23

-2.63 rounding to the second decimal place.

Answer 24

It round's to -2.62

Answer 25

Done

Answer 26

Okay, thanks!

Answer 27

You are welcome

Answer 28

Wait, why is it -2.62? @Pawanyadav

Answer 29

@Pawanyadav

Answer 30

-2.625 if we get 5 after even number
We don't increase it by one
Here 5 is after 2.

Answer 31

I never heard of that rule, I always remembered greater than 5 it goes up 1.

Answer 32

Thats for Greater than 5 not 5

Answer 33

I meant including 5.

Answer 34

you should find that
\[\Large \frac{d^2y}{dx^2} = \frac{-3y^2 - 9x^2}{y^3}\]


Now plug in (x,y) = (1,2)
\[\Large \frac{d^2y}{dx^2} = \frac{-3y^2 - 9x^2}{y^3}\]
\[\Large \frac{d^2y}{dx^2} = \frac{-3(2)^2 - 9(1)^2}{(2)^3}\]
\[\Large \frac{d^2y}{dx^2} = -\frac{21}{8}\]
\[\Large \frac{d^2y}{dx^2} = -2.625\]
\[\Large \frac{d^2y}{dx^2} = -2.63\]

Answer 35

So it is -2.63? Thanks everyone!

Answer 36

That's solve...try to find that rule

Answer 37

I put the answer -2.63 and it said I was right, so I think what I thought was right.

Answer 38

3x^2 + y^2 = 7, differentiating w.r.t x 6x + 2y.y' = 0 or y' = -3x/y differentiating again, 6 + 2y' + 2yy'' = 0 y' = -3x/y = -3/2 putting values, y = 2 and x = 1 y'' = -3/4 = -0.75