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Find dy/dx if x^2y^2 = sin(n+y) .​

User Rrs
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To find dy/dx, we need to use implicit differentiation.

Given the equation: x^2y^2 = sin(n + y)

Step 1: Differentiate both sides of the equation with respect to x.

d/dx(x^2y^2) = d/dx(sin(n + y))

Step 2: Apply the product rule on the left side of the equation.

2x * y^2 + x^2 * 2y(dy/dx) = cos(n + y) * (d/dx(n + y))

Step 3: Simplify the equation.

2xy^2 + 2x^2y(dy/dx) = cos(n + y) * (0 + dy/dx)

Step 4: Solve for dy/dx.

2x^2y(dy/dx) = cos(n + y) * dy/dx - 2xy^2

Step 5: Factor out dy/dx from the right side of the equation.

2x^2y(dy/dx) - cos(n + y) * dy/dx = -2xy^2

Step 6: Factor out dy/dx on the left side of the equation.

dy/dx(2x^2y - cos(n + y)) = -2xy^2

Step 7: Solve for dy/dx.

dy/dx = -2xy^2 / (2x^2y - cos(n + y))

So, the derivative dy/dx of the given equation is -2xy^2 / (2x^2y - cos(n + y)).

User Baris Usanmaz
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