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Question
Find `bb(dy/dx)` for the given function:
xy = `e^((x - y))`
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Solution
Given, xy = `e^((x - y))`
Taking logarithm of both the sides,
log (xy) = ` log e^((x - y))`
log x + log y = (x − y) loge e ....[โต log xy = log x + log y]
log x + log y = x − y ...[โต loge e = 1]
Differentiating both sides with respect to x,
`d/dx log x + d/dx log y = d/dx (x) - d/dx (y)`
`1/x + 1/y dy/dx = 1 - dy/dx `
`1/y dy/dx + dy/dx = 1 - 1/x`
`dy/dx ((1 + y)/y) = 1 - 1/x`
`((1 + y)/y) dy/dx = (x - 1)/x`
`therefore dy/dx = (y (x - 1))/(x (1 + y))`
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