Consider the curve given by the equation 2xy+y^2=16
Find dy/dx
2y + 2xy' + 2yy' = 0
y' = -y/(x+y)
To find dy/dx, we need to differentiate both sides of the equation with respect to x using implicit differentiation.
The equation is 2xy + y^2 = 16.
Differentiating both sides with respect to x, we get:
d(2xy)/dx + d(y^2)/dx = d(16)/dx.
Using the product rule, we can differentiate 2xy with respect to x as follows:
d(2xy)/dx = 2y + 2x(dy/dx).
Differentiating y^2 with respect to x using the chain rule, we have:
d(y^2)/dx = 2y(dy/dx).
Since d(16)/dx = 0 (the derivative of a constant is zero), we can rewrite our equation as:
2y + 2x(dy/dx) + 2y(dy/dx) = 0.
Combining like terms, we have:
2y + 2x(dy/dx) + 2y(dy/dx) = 0.
Simplifying further, we get:
2y + 2x(dy/dx) + 2y(dy/dx) = 0.
Now, we can factor out the common factor of 2(dy/dx) from the terms involving dy/dx:
(2x + 2y)dy/dx + 2y = 0.
Finally, we can solve for dy/dx by isolating it:
(2x + 2y)dy/dx = -2y.
Dividing both sides by (2x + 2y), we get:
dy/dx = -2y / (2x + 2y).
Therefore, the derivative dy/dx of the curve given by the equation 2xy + y^2 = 16 is -2y / (2x + 2y).