Use implicit differentiation to find the second derivative:
2xy^2=4 (nothing is in parenthesis)
Ok, so first you have to realize that 2xy^2 is a u-v problem
So...
U:2x V:y^2
U':2 V':2y(dy/dx)
So dy/dx: 2y^2+2x2y(dy/dx)
dy/ dx = -2y^2/(2x2y)
This then is u-v'd again to get it into the second derivative form
u:-2y^2 v:2x2y(which needs a sep u-v 2 solve)
u':-4y(dy/dx) v':4y+4x(dy/dx)
So the second derivative:
-4y(dy/dx)2x2y+-2y^2(4y+4x(dy/dx))/ (2x2y)^2
Then you plug in answer you got for dy/dx in the second derivative formula, do a bunch of 'fun' algebra and you get the answer (which, by the way is:
8y^3-2y^2(4x)-2y^2(4y)/(2x2y) (which of course, if you feel up to a challege in you orginizational skills, you can simplify
To find the second derivative using implicit differentiation, follow these steps:
Step 1: Differentiate both sides of the equation with respect to x.
The given equation is 2xy^2 = 4.
Differentiating both sides with respect to x, we get:
d/dx (2xy^2) = d/dx (4)
Step 2: Differentiate each term individually.
To differentiate 2xy^2 with respect to x, we consider y as a function of x. Therefore, we need to apply the chain rule.
Applying the chain rule, we get:
d/dx (2xy^2) = 2y^2 * dy/dx + 4xy * d/dx (y)
Step 3: Solve for dy/dx.
Since we're looking for the second derivative, let's solve for dy/dx in the equation 2xy^2 = 4:
2y^2 * dy/dx + 4xy * dy/dx = 0
Rearrange the terms:
dy/dx * (2y^2 + 4xy) = 0
Divide both sides by (2y^2 + 4xy):
dy/dx = 0 / (2y^2 + 4xy) = 0
Step 4: Differentiate dy/dx with respect to x.
Now, differentiate dy/dx with respect to x to find the second derivative, d^2y/dx^2:
(d/dx) (dy/dx) = (d^2y/dx^2) = d/dx (0)
Differentiating 0 with respect to x, we get:
(d^2y/dx^2) = 0
Therefore, the second derivative of the equation 2xy^2 = 4 is 0.