Question

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

Answer 1

Answer:

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