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Question
Find `dy/dx` for the function given in the question:
yx = xy
Solution
Given, yx = xy
Taking logarithm of both sides, log yx = log xy
x log y = y log x
Differentiating both sides with respect to x,
`=> x d/dx log y + log y d/dx (x)`
`= y d/dx log x + log x d/dx y`
`=> x xx 1/y dy/dx + log y xx 1 = y xx 1/x + log x dy/dx`
`=> x/y dy/dx + log y = y/x + log x dy/dx `
`=> x/y dy/dx - log x dy/dx = y /x - log y`
`=> dy/dx (x/y - log x) = y /x - log y`
`=> dy/dx (x^2 - xy log x) = y ^2 - xy log y` On multiplying by xy,
`therefore dy/dx = (y ^2 - xy log y)/(x^2 - xy log x)`
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