find the value of dy/dx when log (xy)=x²+y²
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log(xy) = x² + y²
logx + logy = x² + y². [ log(ab) = log(a) + log(b)]
now, differentiate with respect to x
1/x + 1/y. dy/dx = 2x + 2y.dy/dx [ we know, d(logf(x))/dx = 1/f(x).d(f(x))/dx]
1/x -2x = (2y - 1/y)dy/dx
(1 - 2x²)/x = (2y² -1)/y .dy/dx
(1 - 2x²)y/(2y² -1)x = dy/dx
(y - 2x²y)/(2y²x - x) = dy/dx
dy/dx = (y - 2x²y)/(2y²x - x)
logx + logy = x² + y². [ log(ab) = log(a) + log(b)]
now, differentiate with respect to x
1/x + 1/y. dy/dx = 2x + 2y.dy/dx [ we know, d(logf(x))/dx = 1/f(x).d(f(x))/dx]
1/x -2x = (2y - 1/y)dy/dx
(1 - 2x²)/x = (2y² -1)/y .dy/dx
(1 - 2x²)y/(2y² -1)x = dy/dx
(y - 2x²y)/(2y²x - x) = dy/dx
dy/dx = (y - 2x²y)/(2y²x - x)
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Answer:y(1-2x²)/x(2y²-1)
Step-by-step explanation:
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