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प्रश्न
If ex + ey = ex+y, then show that `"dy"/"dx" = -e^(y - x)`.
उत्तर
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^y"dy"/"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)` ...[By (1)]
= `e^y/-e^x`
= – ey–x.
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