Given that x
2
y
2 = cos
x
2
y
2
�
,
determine dy
dx
Taking the derivative of both sides with respect to x using the chain rule, we get:
2x(2y)(dy/dx) - 2y(2x)(sinx)(dx/dx) = 0
Simplifying, we get:
2xy(dy/dx) - 2xy(sin x) = 0
Dividing both sides by 2xy, we get:
dy/dx - sin x = 0
Therefore:
dy/dx = sin x
To find dy/dx, we need to apply the implicit differentiation technique. The given equation is:
x^2 * y^2 = cos(x^2 * y^2)
Now, we differentiate both sides of the equation with respect to x:
d/dx (x^2 * y^2) = d/dx (cos(x^2 * y^2))
Using the product rule on the left side:
(2x * y^2) + (x^2 * 2y * dy/dx) = d/dx (cos(x^2 * y^2))
Applying the chain rule on the right side:
(2x * y^2) + (x^2 * 2y * dy/dx) = -sin(x^2 * y^2) * d/dx (x^2 * y^2)
Using the chain rule again:
(2x * y^2) + (x^2 * 2y * dy/dx) = -sin(x^2 * y^2) * (2x * y^2 + 2y^2 * dy/dx)
Now, let's solve for dy/dx by moving all the terms involving dy/dx to one side:
x^2 * 2y * dy/dx + sin(x^2 * y^2) * 2y^2 * dy/dx = -(2x * y^2) + sin(x^2 * y^2) * (2x * y^2)
Factoring out dy/dx:
dy/dx * (x^2 * 2y + sin(x^2 * y^2) * 2y^2) = -(2x * y^2) + sin(x^2 * y^2) * (2x * y^2)
Finally, divide both sides by (x^2 * 2y + sin(x^2 * y^2) * 2y^2) to obtain dy/dx:
dy/dx = (-(2x * y^2) + sin(x^2 * y^2) * (2x * y^2)) / (x^2 * 2y + sin(x^2 * y^2) * 2y^2)
So, the value of dy/dx is given by the above expression.