If 3x+xy+4y=8, what is the value of d^2y/dx^2 at the point (1,1)? My work:

3+(x)(dy/dx)+(y)+4(dy/dx)=0
dy/dx(x+4)=-3-y
dy/dx=-3-y/x+4

2nd Dervative: (x+4)(0-dy/dx)-(-3-y)/(x+4)^2
(x+4)(-(-y-3/x+4))-(-3-y)/(x+4)^2
(x+4)(y+3)/(-x+4)+(y+3)/(x+4)^2
*Mult. everything by (-x+4)
Simplify:

(x+4)(y+3)+(y+3)(-x+4)/(x+4)^2(-x+4)
*The (-x+4) cancels

Left over: (x+4)(y+3)+(y+3)/(x+4)^2
*Plug in (1,1)

3x+xy+4y=8

3 + y + xy' + 4y' = 0
y'(x+4) = -(y+3)

y' = -(y+3)/(x+4)

y" = -[(y')(x+4)-(y+3)(1)]/(x+4)^2
= -[-(y+3)/(x+4)*(x+4) - (y+3)]/(x+4)^2
= -(-3-y-y-3)/(x+4)^2
= (2y+6)/(x+4)^2

check:
3x+xy+4y=8
y = (8-3x)/(x+4)
y' = -20/(x+4)^2
y" = 40/(x+4)^3

Since y = (8-3x)/(x+4), that means
2y+6 = (16-6x + 6(x+4))/(x+4)
= (16-6x+6x+24)/(x+4)
= 40/(x+4)
So,
y" = (2y+6)/(x+4)^2
= 40/(x+4)^3

So the implicit and explicit vales agree.

To find the value of d^2y/dx^2 at the point (1,1) using the given equation, we first need to find dy/dx.

Given equation: 3x + xy + 4y = 8

Rearranging for dy/dx:
3 + x(dy/dx) + y + 4(dy/dx) = 0
(dy/dx)(x + 4) = -3 - y
dy/dx = (-3 - y) / (x + 4)

Now, let's differentiate the expression for dy/dx with respect to x to get the second derivative d^2y/dx^2.

d^2y/dx^2 = [(x + 4)(0 - dy/dx) - (-3 - y)] / (x + 4)^2
= (x + 4)(-(-y - 3)/(x + 4)) / (x + 4)^2 + (-3 - y) / (x + 4)^2

Simplifying further:
d^2y/dx^2 = (x + 4)(y + 3) / (-x + 4) + (y + 3) / (x + 4)^2

To compute the value of d^2y/dx^2 at the point (1,1), substitute x = 1 and y = 1 into the expression:

d^2y/dx^2 = (1 + 4)(1 + 3) / (-1 + 4) + (1 + 3) / (1 + 4)^2
= (5)(4) / 3 + 4 / 25
= 20/3 + 4/25
= 505/75
= 6.73

Therefore, the value of d^2y/dx^2 at the point (1,1) is approximately 6.73.