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)

Hi I know how to do this problem, but can you see if i did it right?

Sure! Let's check your work step by step.

Starting with the equation 3x + xy + 4y = 8, you correctly wrote the equation in terms of derivatives:

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

Then, you isolated dy/dx:

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

So far, everything looks good.

Now, let's calculate the second derivative. You correctly used the quotient rule:

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

Then, you simplified the expression:

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

Next, you multiplied everything by (-x + 4) to simplify further, which is a correct step.

After multiplying, you obtained:

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

However, there seems to be a small mistake when you canceled out (-x + 4) in the denominator. It should be canceled out with one of the factors in the numerator, not just completely canceled.

The correct simplification would be:

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

Finally, you mentioned plugging in the point (1, 1). To find the value of (d^2y/dx^2) at that point, substitute x = 1 and y = 1 into the expression.

(d^2y/dx^2) = [(1 + 4)(1 + 3) + (1 + 3)(-1 + 4)]/[(1 + 4)^2(-1 + 4)]

Simplifying further would give you the final answer at the point (1, 1).

I hope this clarifies your work and helps you verify if you did it correctly. Let me know if you have any more questions!