2x^2+y^2=8xy find dy/dx

2 (x-2y) / (4x-y) = dy/dx

Step-by-step explanation:

2x^2+y^2=8xy

Take the derivative of each term, remembering that we take the 8xy as derivative by parts)

2 * 2x dx + * 2y dy = 8 (x dy + dx * y)

4x dx + 2y dy = 8x dy + 8y dx

Subtract 2y dy from each side

4x dx + 2y dy - 4y dy = 8x dy - 2y dy + 8y dx

4x dx = 8x dy - 2y dy + 8y dx

Subtract 8y dx from each side

4x dx - 8y dx = 8x dy - 2y dy + 8y dx-8y dx

4x dx - 8y dx = 8x dy - 2y dy

(4x-8y) dx = (8x-2y) dy

Factor out a 4 from the left side and a 2 from the right side

4 (x-2y) dx = 2 (4x-y) dy

Cancel a 2

2 (x-2y) dx = (4x-y) dy

Divide each side by (4x-y)

2 (x-2y) / (4x-y) dx = (4x-y) / (4x-y) dy

2 (x-2y) / (4x-y) dx = dy

Divide by dx

2 (x-2y) / (4x-y) dx/dx = dy/dx

2 (x-2y) / (4x-y) = dy/dx