Question

log(xy)=x2+y2 find dy/dx

Answer 1

Answer:

-y/x (2x²-1)/(2y²-1)

Step-by-step explanation:

log(xy)= x²+y²

x²+y²=log(xy)

to find dy/dx

2x+2ydy/dx = 1/(xy) { xd/dx y + y d/dx x}

2x+2ydy/dx = 1/(xy) {xdy/dx +y}

2x+2ydy/dx = 1/y.dy/dx +1/x

2ydy/dx -1/ydy/dx = 1/x-2x

dy/dx (2y²-1)/y = (1-2x²)/x

dy/dx = y/x (1-2x²)/(2y²-1)

or

dy/dx = -y/x (2x²-1)/(2y²-1) Ans..