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Question
If xy . yx , then prove that `"dy"/"dx" = y/x((x log y - y)/(y log x - x))`
Solution
Given xy . yx
Taking logarithm on both sides we get,
y log x = x log y
Differentiating with respect to 'x' we get,
`y*1/x + log x * "dy"/"dx" = x * 1/y "dy"/"dx" + log y(1)`
`=> y/x + log x ("dy"/"dx") = x/y "dy"/"dx" + log y`
`=> "dy"/"dx" (log x - x/y) = log y - y/x`
`=> "dy"/"dx" ((y log x - x)/y) = (x log y - y)/x`
`=> "dy"/"dx" = (x log y - y)/x xx y/(y log x - x)`
`=> "dy"/"dx" = y/x((x log y - y)/(y log x - x))`
Hence proved.
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