find dy/dx implicit differentiation
x^2y+y^2x=0
at x=-2, x=2
2xydx+x^2 dy + 2yxdy+y^2 dx=0
dy/dx(x^2+2yx)=-(y^2/2xy)
ok, at the given point, you can figure dy/dx. I am not certain what you mean for the point to be.
x^2(dy/dx) + 2xy + y^2 + 2xy(dy/dx) = 0
(dy/dx)(x^2 + 2xy) = - y^2 - 2xy
dy/dx = (-y^2-2xy)/(x^2 + 2xy)
when x = -2 in original
4y + 2y^2 = 0
2y(2 + y)=0
y = 0 or y = -2
now plug in (-2,0) and (-2,-2) into dy/dx
do the same using x = 2
Do you mean x = -2, y = 2?
Do you mean x^2*y + y^2*x = 0 ?
What are the exponents?
2 or 2x and 2y?
x cannot have two values unless you want the derivative at two different points.
Anyway, implicit differentiation with respect to x results in:
x^2*dy/dx + y*2x + y^2 + 2y = 0
I messed up in the calculation of the point
when x = -2
4y - 2y^2 = 0
2y(2-y)=0
y = 0 or y = 2
So the second point is (-2,2) not (-2,-2)
To find dy/dx using implicit differentiation, we need to differentiate both sides of the equation with respect to x.
Let's differentiate the equation x^2y + y^2x = 0 term by term:
The derivative of x^2y with respect to x is 2xy + x^2(dy/dx), applying the product rule.
The derivative of y^2x with respect to x is y^2 + 2xy(dy/dx), again applying the product rule.
The derivative of 0 with respect to x is 0.
Now we can rewrite the differentiated equation:
2xy + x^2(dy/dx) + y^2 + 2xy(dy/dx) = 0.
Combine like terms:
2xy + 2xy(dy/dx) + x^2(dy/dx) + y^2 = 0.
Factor out dy/dx:
(2xy + x^2 + 2xy)(dy/dx) = -2xy - y^2.
Simplify:
(4xy + x^2)(dy/dx) = -2xy - y^2.
Finally, solve for dy/dx:
dy/dx = (-2xy - y^2) / (4xy + x^2).
To find values of dy/dx at x = -2 and x = 2, simply substitute these values into the expression for dy/dx:
For x = -2:
dy/dx = (-2(-2)y - y^2) / (4(-2)y + (-2)^2)
= (4y - y^2) / (-8y + 4)
For x = 2:
dy/dx = (-2(2)y - y^2) / (4(2)y + 2^2)
= (-4y - y^2) / (8y + 4)