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