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Question
If ex + ey = e(x + y), then show that `dy/dx = -e^(y - x)`.
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Solution
ex + ey = ex + y ...[1]
Differentiating both sides w.r.t. x, we get
`e^x + e^y. dy/dx = e^(x + y).d/dx(x + y)`
∴ `e^x + e^y.dy/dx = e^(x + y).(1 + dy/dx)`
∴ `e^x + e^ydy/dx = e^(x + y) + e^(x + y)dy/dx`
∴ `(e^y - e^(x + y))dy/dx = e^(x + y) – e^x`
∴ `dy/dx = (e^(x +y) - e^x)/(e^y - e^(x + y)`
= `((e^x + e^y) - e^x)/(e^y - (e^x + e^y))` ...[Using (1), since `e^(x + y) = e^x + e^y`]
= `e^y/-e^x`
= `dy/dx − e^(y − x)`
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