Find d2y/dx2 in terms of x and y.
x^2y^2-2x=3
d/dx (x^2y^2 - 2 x)
= 2 x y^2 - 2
d/dx (2 x y^2 - 2)
= 2 y^2
x^2y^2 - 2x = 3
2xy^2 + 2x^2yy' - 2 = 0
y' = (2-2xy^2)/2x^2y = (1-xy^2)/x^2y
Now use the quotient rule to get
y" = ((-y^2 - 2xyy')(x^2y) - (1-xy^2)(2xy + x^2y')) / x^4y^2
= (2x^4-2xy^2-1) / x^4y^3
Now, if you wanted partial derivatives, go with Damon.
To find d2y/dx2 (the second derivative of y with respect to x), we need to first find dy/dx (the first derivative of y with respect to x).
Let's differentiate the given equation implicitly with respect to x.
Differentiating the equation x^2y^2 - 2x = 3 with respect to x, we get:
2xy^2 + x^2(2y)(dy/dx) - 2 = 0
Rearranging the equation, we have:
x^2(2y)(dy/dx) = 2 - 2xy^2
Now, solving for dy/dx, we get:
dy/dx = (2 - 2xy^2) / (2xy^2)
Next, we need to differentiate dy/dx with respect to x to find d2y/dx2.
Using the quotient rule, we differentiate the numerator and denominator separately and then apply the formula.
For the numerator:
d/dx (2 - 2xy^2) = -2y^2 - 2x(2y)(dy/dx)
For the denominator:
d/dx (2xy^2) = 2y^2(dy/dx) + 2xy(2y)(dy/dx)
Now we can substitute these expressions back into the formula:
d2y/dx2 = ( (-2y^2 - 2x(2y)(dy/dx)) * (2xy^2) - (2y^2(dy/dx) + 2xy(2y)(dy/dx)) * (2 - 2xy^2) ) / (2xy^2)^2
Simplifying this expression will give us the second derivative of y with respect to x in terms of x and y.