d/dx (x^2y^2 - 2 x)
= 2 x y^2 - 2
d/dx (2 x y^2 - 2)
= 2 y^2
Find d2y/dx2 in terms of x and y.
x^2y^2-2x=3
2 answers
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.
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.
1 answer