Question

If x^2+3y^2=8+2xy, then d^2y/dx^2 at (2,2) =

Answer 1

We are given
x² + 3y² = 8 + 2xy
We are asked for
d^2y/dx^2 at (2,2)
Which is the second derivative of the expression in terms of x evaluated at (2,2)

The expression can't be express explicitly in terms of x
Doing implicit differentiation
2x dx + 6y dy = 2y dx + 2 x dy
Solving for dy/dx
dy/dx = (2x - 2y) / (2x - 6y)
Then, 
d^2y/dx^2 at (2,2) can now be determined

Answer 2

Answer:

The value of $\frac{d^2y}{dx^2}$ at (2,2) is $\frac{-1}{4}$.

Step-by-step explanation:

The given equation is

$x^2+3y^2=8+2xy$

Differentiate with respect to x.

$2x+3(2yy')=0+2(xy'+y)$

Here, $y'=\frac{dy}{dx}$

$2(x+3yy')=2(xy'+y)$

$x+3yy'=xy'+y$

$3yy'-xy'=y-x$

$(3y-x)y'=y-x$

$y'=\frac{y-x}{3y-x}$

The value of first derivative at (2,2) is

$y'_{(2,2)}=\frac{2-2}{3(2)-2}=0$

Differentiate with respect to x. Using quotient rule we get,

$y''=\frac{(3y-x)(y'-1)-(y-x)(3y'-1)}{(3y-x)^2}$

The value of second derivative at (0,0) is

$y''_{(2,2)}=\frac{(3(2)-2)(y'_{(2,2)}-1)-(2-2)(3y'_{(2,2)}-1)}{(3(2)-(2))^2}$

$y''_{(2,2)}=\frac{(4)(0-1)-0}{4^2}$

$y''_{(2,2)}=\frac{-4}{16}$

$y''_{(2,2)}=\frac{-1}{4}$

Therefore the value of $\frac{d^2y}{dx^2}$ at (2,2) is $\frac{-1}{4}$.